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Honors Chemistry

Topic 1: An Introduction to Chemistry Matter & Change (Chapter 1 in Modern Chemistry)

Chemistry is a physical science. It is the study of the composition, structure, and properties of matter, the processes that matter undergoes, and the energy changes that accompany these processes.

Branches of Chemistry

(6 main areas of study)

x Organic chemistry-the study of most carbon-containing compounds (except carbon & carbonate)

x Inorganic chemistry-the study of non-organic substances x Physical chemistry-the study of the properties and changes of matter and their relation to

energy x Analytical chemistry-the identification of the components and composition of materials x Biochemistry-the study of substances and processes occurring in living things x Theoretical chemistry-the use of mathematics and computers to understand the principles

behind observed chemical behavior and to design and predict the properties of new compounds

A chemical is any substance that has a definite composition.

Mass is a measure of the amount of matter.

Matter is anything that has mass and takes up space. Matter is stuff.

Matter has characteristic properties. These properties can be used to distinguish among substances and to separate them. Properties can be classified as physical or chemical.

Physical properties are characteristic that can be observed or measured without changing the identity of the substance. It may look different but it has the same makeup. For example: boiling, melting, and freezing points are physical properties. Water has a formula of H2O when it boils and becomes steam which also has a formula of H2O. It has this same formula when liquid water freezes and changes to ice. Physical changes are changes in a substance that does not involve a change in the identity of the substance. All phase changes (changes of state) are physical changes.

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Matter in the solid state has definite volume and definite shape or according to the Kinetic Molecular Theory (KMT), solid particles vibrate around fixed points. Matter in the liquid state has a definite volume but does not have a definite shape. It assumes the shape of its container. According to the (KMT), liquid particles vibrate around moving points. Matter in the gas state has neither definite volume nor definite shape or according to the (KMT), gas particles move in random straight line motion until they hit something. A fourth state of matter is plasma. Plasma is a high-temperature physical state of matter in which atoms lose most of their electrons.

http://www.dlt.ncssm.edu/tiger/Flash/phase/KineticEnergy-Solid.html http://www.dlt.ncssm.edu/tiger/Flash/phase/KineticEnergy-Liquid.html http://www.dlt.ncssm.edu/tiger/Flash/phase/KineticEnergy-Gas.html Physical properties can be intensive or extensive. Extensive properties depend on the amount of matter that is present. For example: volume, mass, and the amount of energy in a substance Intensive properties do not depend on the amount of matter present. For example: melting point, boiling point, density, and ability to conduct electricity and to transfer energy as heat

Chemical properties relate to a substance’s ability to undergo changes that transform it into different substances. The substances react to form new substances. A chemical change or chemical reaction is a change in which one or more substances are converted into different substances. The substances that react in a chemical change are called the reactants. The substances that are formed by the chemical change are called the products.

carbon plus oxygen yield carbon dioxide.

carbon + oxygen Æ carbon dioxide

C + O2 Æ CO2

Task 1a

1. Label the following as chemical or physical change/property. a. Melting b. Digestion c. Rusting of iron d. Tearing paper

2. Label the following as an intensive or extensive physical property. a. Length b. Mass c. Density d. Color

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Classification of Matter

Matter is broken down into mixtures and pure substances.

Mixtures are blends of two or more kinds of matter, each of which retains its own identity and properties. They are simply physically mixed together and have variable composition. In other words, they do not have a formula. They can usually be separated relatively easily. There are heterogeneous mixtures and homogeneous mixtures. A heterogeneous mixture is not mixed uniformly, for example: granite, clay & water. Homogeneous mixtures are uniform in composition. They still do not have a formula, but the particles are physically mixed evenly all the way through. Homogeneous mixtures are also called solutions. Examples of solutions are salt water, air, and kool-aid. Some mixtures can be separated by filtration, evaporation, distillation, or chromatography.

Pure Substances are all homogeneous but are not mixtures because they are chemically combined instead of physically combined. A pure substance has a fixed composition (has a formula). Every sample of a given pure substance has exactly the same characteristic properties and has exactly the same composition. Pure substances can be broken down into elements and compounds.

Elements are pure substances that cannot be broken down into simpler, stable substances and is made of one type of atom. Elements are found on the periodic table. They are made of only one type of atom. An atom is the smallest unit of an element that maintains the chemical identity of that element.

A compound is a substance that can be broken down into simple stable substances. Each compound is made from the atoms of two or more elements that are chemically bonded. They have definite formulas.

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Task 1b

1. Label the following as heterogeneous mixture, solution, compound or element. a. Salt, NaCl b. Tea c. Iced tea d. Iron e. Concrete f. Copper g. Sugar

The Periodic Table

The periodic table is one of the most important references for chemists. It is imperative that you learn the symbols and names of common elements. Be sure to study the list supplied in the memory work section of edline.

Periodic table key: blue main group metals pink transition metals teal metalloids purple nonmetals green noble gases

Symbol color key (at rt): Black = solid White = gas Yellow = liquid

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The vertical columns on the periodic table are called groups or families. We refer to the groups by number and the families by name. For example, column 2 is called group 2 or the alkaline earth metal family. Elements in the same group have similar properties. There are a few common family names that you need to know. Other families are known by the first element in that group.

Group 1 is the alkali metals family Group 2 is the alkaline earth metal family Group 17 is the halogen family Group 18 is the noble gas family The horizontal rows of elements in the periodic table are called periods. Physical and chemical properties change somewhat regularly across a period. Periods are designated by the number of row (1-7). Elements can also be classified as metals, nonmetals, and metalloids. Metals are to the left of a stair-step line that begins on the left of Boron (except for Hydrogen which is a nonmetal). Nonmetals are to the right of the stair-step line. Elements that touch the stair-step line on a top, bottom or side (not corner) except for aluminum are metalloids. Metals are good conductors of heat and electricity. They are malleable and ductile. They tend to lose valence electrons when they bond. Nonmetals are poor conductors of heat and electricity. They are gaseous or brittle. They tend to gain electrons when they bond. Metalloids sometimes act like metals and sometimes act like nonmetals. (See color codes on periodic table) Task 1c

1. Write the symbol for the following elements. a. Copper b. Sodium c. Argon d. Oxygen e. Zinc

2. Write the name of the following element symbols. a. K b. Li c. S d. He e. Fe

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3. List the group and period of the following elements. a. Xe b. Cr c. I d. U

4. Label the following as a metal, nonmetal, or metalloid. a. Aluminum b. Nickel c. Germanium d. Carbon

Measurements & Calculations (Chapter 2 in Modern Chemistry)

The Scientific Method is a logical approach to solving problems by observing and collecting data, formulating hypotheses, testing hypotheses, and formulating theories that are supported by data. Observing uses the senses. Observations can be qualitative or quantitative. Qualitative observations are non-numerical, for example: the room is cold. Quantitative observations are numerical, for example: the room is 58oF.

Chemists study systems. A system is a specific portion of matter in a given region of space that has been selected for study during an experiment or observation. An example of a system is a test tube and its contents.

An hypothesis is a testable statement (educated guess). Many times hypotheses are stated as an “if-then” statement.

Hypotheses are tested using an experiment. For an experiment to be valid, the experiment will have controls or conditions that are constant throughout the testing. The experiment will also have variables, or conditions that change. Experiments will usually have a dependent variable and an independent variable. The independent variable (usually on the x axis of a graph) is the variable that is typically being manipulated by the experimenter while the dependent variable (usually on the y axis of a graph) is the observed result of the independent variable being manipulated. An example would be the amount of fertilizer vs. plant growth. The amount of fertilizer would be the independent variable and the growth would be the dependent variable.

After the hypothesis has been tested and the data has been recorded and analyzed, then conclusions can be drawn from the experiment. Theorizing about the experiment could start by constructing a model. A model in science is more than a physical object; it is often an explanation of how phenomena occur and how data or events are related (For example: the

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atomic model). If a model successfully explains many phenomena, it becomes part of a theory. A theory is a broad generalization that explains a body of facts or phenomena. A scientific theory summarizes a hypothesis or group of hypotheses that have been supported with repeated testing. A theory is valid as long as there is no evidence to dispute it. Therefore, theories can be disproven. This is different from a law. A law generalizes a body of observations. At the time it is made, no exceptions have been found to a law. Scientific laws explain things, but they do not describe them. One way to tell a law and a theory apart is to ask if the description gives you a means to explain 'why'.

Task 1d

1. Label the following as qualitative or quantitative observations. a. The liquid floats on water. b. The metal is malleable. c. The liquid has a temperature of 55.6oC.

SI Measurements & Units

Quantity is something that has magnitude, size, or amount. A measurement is a quantity with a unit of measurement. For example: 1.5 g. Scientists us the SI system of measurement (Le Systeme International d’Unites). We will also refer to this as the metric system. Here is a table of the base (fundamental) units used in the SI system. We will only be using the first five.

Important fundamental measurements

Mass- a measure of the quantity of matter, measured with a balance, units g, kg, etc.

Length- the distance between two points, measure with a ruler, units m, km, cm, mm, etc.

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Derived SI Units

Units not listed on the base unit table are derived units. Derived units are combinations of SI base units.

Important derived measurements

Weight- a measure of the gravitational pull on matter, measured with a scale, unit N (newtons).

Volume- the amount of space occupied by an object, measures with a ruler or graduated cylinder, units m3, cm3, mL, L, etc. For regular shape objects, you can use a mathematical formula for volume such as V = l x w x h IMPORTANT: 1 cm3 = 1 mL and 1 dm3 = 1 L.

Density- the ratio of mass to volume, or mass divided by volume. D = m/V, the units for density are g/cm3, g/mL, kg/L, kg/dm3, etc.

D = m V

Task 1e 1. Label each of the following measurements by the quantity each represents. For instance,

a measurement of 10.6 kg/m3 represents density. a. 5.0 g/mL b. 37 s c. 39.56 g d. 47 J e. 25.3 cm3 f. 500 m2 g. 30.23 mL

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Task 1f 1. What is the density of a block of marble that occupies 310. mL and has a mass of 853 g? 2. Diamond has a density of 3.26 g/cm3. What is the mass of a diamond that has a volujme

of 0.351 cm3? 3. What is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that the

density of mercury is 13.6 g/mL? 4. A block of sodium that has the measurements 3.00 cm x 5.00 cm x 5.00 cm has a mass of

75.5 g. Calculate the density of sodium.

Dimensional Analysis (Factor Label Method)

Dimensional Analysis is a mathematical technique that allows you to use units to solve problems involving measurements. THIS IS AN IMPORTANT TECHNIQUE!

Use conversion factors to go from the quantity given to the quantity sought. Factors are numbers, labels are units.

When using the factor-label method, problems consist of three parts: 1. a known beginning – GIVEN 2. a desired end – WANTED 3. a connecting path – CONVERSION FACTORS

A conversion factor is a ratio derived from the equality between two different units that can be used to convert from one unit to the other. For example:

4 quarters = 1 dollar

This can be written as two conversion factors:

4 quarters or 1 dollar 1 dollar 4 quarters

Notice that each conversion factor equals 1. They equal each other. Notice conversion factors can be flipped.

Here is an example using conversion factors & dimensional analysis: How many seconds are in 2.5 days?

2.5 days 24 hours 60 minutes 60 seconds 1 day 1 hour 1 minute

Multiply the top, multiply the bottom, and divide the answers. All units cancel out except seconds.

This equals 216 000 seconds.

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Metric Conversions

Learn this chart! (Here’s an easy way: My kangaroo has dance until dawn ‘cause music makes noise).

106 103 102 101 100 10-1 10-2 10-3 10-6 10-9

M k h da u d c m P n

Mega- kilo- hecta- deka- deci- centi- milli- micro- nano-

You can use dimensional analysis with this information or move the decimal the appropriate number of places. For example:

Convert 35 m to km. 35 m 1 km = 0.035 km 1000 m Or “m” has no prefix so it is in the units column, “km” has kilo as its prefix, so it is in the “k” column. Move the decimal 3 places to the left, just like on the chart. If you use this method, watch out for prefixes that change by more than one change of 10 (mega, micro, nano).

35 m becomes 0.035 km

Task 1g

1. Complete the following conversions a. 10.5 g = _____ kg b. 1.57 km = _____ m c. 3.54 Pg = _____ g d. 3.5 mol = _____Pmol e. 1.2 L = _____ mL f. 37.8 mL = _____ cm3

2. Use dimensional analysis to determine the following.

a. If 1 mag = 13 bops and 1 bop = 4.6 skuts, how many mags are in 583 skuts? b. How many centimeters are in 2.5 yards? (1 inch = 2.54 cm)

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Temperature Conversions

Use the following formulas for temperature conversions.

K = oC + 273 oC = K – 273

oC = 5/9(oF-32) oF = 9/5 oC +32

Task 1h

1. Make the following temperature conversions. a. 58.7oC = _____ K b. 323 K = _____ oC c. 75oF = _____ oC d. 368 K = _____ oF

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Using Scientific Measurements

Accuracy refers to the closeness of measurements to the correct or accepted value of the quantity measured. Precision refers to the closeness of a set of measurements of the same quantity made in the same way.

Percentage Error is calculated by subtracting the accepted value from the experimental value, dividing the difference by the accepted value, and then multiplying by 100.

Percentage error = experimental - accepted x 100 accepted

Task 1i

1. What is the percentage error for a mass measurement of 17.7 g, given that the correct value is 21.1 g?

2. A volume is measured experimentally as 4.26 mL. What is the percent error, given that the correct value is 4.15 mL?

3. During an experiment, a student obtains the following density data: 9.12 g/mL, 9.11 g/mL, and 9.13 g/mL. The literature value for the density of this substance is 8.78 g/mL. Is this student accurate? Is this student precise? What is the % error of this students information?

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Significant figures (Significant Digits) in a measurement consist of all the digits known with certainty plus one final digit, which is somewhat uncertain or is estimated.

Rules for Significant Figures Read from the left and start counting sig figs when you encounter the first non-zero digit

1. All non zero numbers are significant (meaning they count as sig figs) 613 has three sig figs 123456 has six sig figs

2. Zeros located between non-zero digits are significant (they count) 5004 has four sig figs 602 has three sig figs 6000000000000002 has 16 sig figs!

3. Trailing zeros (those at the end) are significant only if the number contains a decimal point; otherwise they are insignificant (they don’t count) 5.640 has four sig figs 120000. has six sig figs 120000 has two sig figs – unless you’re given additional information in the

problem

4. Zeros to left of the first nonzero digit are insignificant (they don’t count); they are only placeholders! 0.000456 has three sig figs 0.052 has two sig figs 0.000000000000000000000000000000000052 also has two sig figs!

Rules for addition/subtraction problems Your calculated value cannot be more precise than the least precise quantity used in the calculation. The least precise quantity has the fewest digits to the right of the decimal point. Your calculated value will have the same number of digits to the right of the decimal point as that of the least precise quantity. In practice, find the quantity with the fewest digits to the right of the decimal point. In the example below, this would be 11.1 (this is the least precise quantity).

7.939 + 6.26 + 11.1 = 25.299 (this is what your calculator spits out)

In this case, your final answer is limited to one sig fig to the right of the decimal or 25.3 (rounded up).

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Rules for multiplication/division problems The number of sig figs in the final calculated value will be the same as that of the quantity with the fewest number of sig figs used in the calculation. In practice, find the quantity with the fewest number of sig figs. In the example below, the quantity with the fewest number of sig figs is 27.2 (three sig figs). Your final answer is therefore limited to three sig figs.

(27.2 x 15.63) 1.846 = 230.3011918 (this is what you calculator spits out)

In this case, since your final answer it limited to three sig figs, the answer is 230. (rounded down) Rules for combined addition/subtraction and multiplication/division problems First apply the rules for addition/subtraction (determine the number of sig figs for that step), then apply the rules for multiplication/division. Task 1j

1. Provide the number of sig figs in each of the following numbers: (a) 0.0000055 g _____ (b) 3.40 x 103 mL ______ (c) 1.6402 g _____ (d) 1.020 L _____ (e) 16402 g ______ (f) 1020 L _______

2. Perform the operation and report the answer with the correct number of sig figs.

(a) (10.3 m) x (0.01345 m) = ___________________ (b) (10.3) + (0.01345) = ______________________ (c) [(10.3) + (0.01345)] = ____________________________

[(10.3) x (0.01345)]

3. Polycarbonate plastic has a density of 1.2 g/cm3. A photo frame is constructed from two 3.0 mm sheets of polycarbonate. Each sheet measures 28 cm by 22 cm. What is the mass of the photo frame?

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Scientific Notation (You should already know this)

Scientific Notation was developed in order to easily represent numbers that are either very large or very small. Scientific Notation is based on powers of the base number 10.

Examples: The number 200,000,000,000 stars in scientific notation is written as 2 x 1011 stars

The number 0. 000,006,645 kilograms in scientific notation is written as 6.645 x 10-6 stars

The first number 6.645 is called the coefficient. The coefficient must be greater than or equal to 1 and less than 10.

The coefficient contains only significant digits. The second number is called the base. The base must always be 10 in scientific notation. The number -6 is referred to as the exponent or power of ten. The exponent must show the number of places that the decimal needs to be moved to change the number to standard notation. A negative exponent means that the number written in standard notation is less than one. To Change from Standard Form to Scientific Notation:

1. Place decimal point such that there is one non-zero digit to the left of the decimal point. 2. Count number of decimal places the decimal has "moved" from the original number.

This will be the exponent of the 10. 3. If the original number was less than 1, the exponent is negative; if the original number

was greater than 1, the exponent is positive.

To Change from Scientific Notation to Standard Form:

1. Determine the number of places the decimal must be moved from the exponent. 2. Decide if the standard form will be a number greater than one or less than one. 3. Move the decimal in the coefficient adding place holders if necessary.

Task 1k

1. Write the following numbers in scientific notation. a. 96 400 b. 0.361 c. 0.0057300 d. 6 587 234 000

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2. Write the following numbers in standard notation. a. 3.97 x 103 b. 8.862 x 10-1 c. 6.251 x 109 d. 512 x 10-8

Data Manipulation

All measurements taken in lab must be in the correct significant digits. Data tables should be prepared ahead of time if possible. You will be expected to understand what is occurring, not only following directions. This is not just cookbook chemistry.

You will have to graph data that you collect in lab. Remember that the independent variable goes on the x-axis. The dependent variable goes on the y-axis.

Two quantities are directly proportional to each other if dividing one by the other gives a constant value. y/x = k or y = kx (in the form of y = mx + b) This is a straight line graph. An example of this is density.

Two quantities are inversely proportional to each other if their product is constant. xy = k This graph produces a curve called a hyperbola. Pressure & volume of gases give this type of inverse proportion.

States of Matter (Chapter 10 Modern Chemistry)

Watch this!! This will review the six phase changes that matter undergoes.

http://www.kentchemistry.com/links/Matter/PhaseChanges.htm

Phase Diagrams

A phase diagram is a graph of pressure versus temperature that shows the conditions under which the phases of a substance exist.

The triple point of a substance indicates the temperature and pressure conditions at which the solid, liquid, and vapor of the substance can coexist at equilibrium.

The critical point of a substance indicates the critical temperature and critical pressure. This is the point above which the substance cannot exist in the liquid state.

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The normal melting point or boiling point is the melting or boiling point at standard pressure. In order to find the normal points you must read the graph across from 1 atm, 760 mmHg, or 101.3 kPa.

General Phase Diagram

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Task 1L

1. Using phase Diagram for water (a), what is the critical point of water? 2. What is water’s triple point? 3. Using the phase diagram for carbon dioxide (b), what is the normal melting point? The

normal boiling point?

Heating & Cooling Curves

Heating & Cooling curves can help you tell at what temperature a substance melts/freezes or boils/condenses at the current pressure. Note that as a phase is change the temperature doesn’t change. All the energy is going toward breaking the particles intermolecular bonds so the change can occur. After the phase change occurs the energy can now be used to increase the temperature (or vice versa).

http://www.dlt.ncssm.edu/tiger/Flash/phase/HeatingCurve.html

This link allows you to perform three melting/boiling experiments while simultaneously graphing the data. You should be able to determine the melting point and boiling point of each substance.

http://www.kentchemistry.com/links/Matter/PhaseChanges.htm

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Solutions (Chapter 12 Modern Chemistry)

Solubility & Solubility Curves

Solubility of a substance is the amount of that substance required to form a saturated solution with a specific amount of solvent at a specified temperature.

A saturated solution contains the maximum amount of dissolved solute. An unsaturated solution contains less solute than a saturated solution under the existing conditions. A supersaturated solution contains more dissolved solute than a saturated solution contains under the same conditions.

As you can see in the diagram below, the solubility of solids generally increase with temperature, while the solubility of gases decrease with a temperature increase.

Task 1m

4. Using the solubility curves, what is the solubility of KClO3 in 100 g of H2O at 50 oC? 5. How many grams of KCl is needed to make a saturated solution at 50 oC in 100 g of

H2O? What about in 200 g of H2O at this same temperature? 6. How many grams of NaCl will dissolve in 100 g of H2O at 60 oC? How much salt would

sink to the bottom of the beaker if 30 g of NaCl is added to 100 g of H2O? What is 60 g of NaCl is added to 100 g of H2O?

7. What type of solution would 40 g of NH3 in 100 g H2O at 20 oC?

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Topic 2: Atomic Theory & Structure Atoms: The Building Blocks of Matter

(Chapter 3 in Modern Chemistry)

History of Atom Theory

Democritus a great Greek thinker, (460-370 BC) first proposed the existence of an ultimate particle. He used the word "atomos" to describe this particle. He also explain solids by stating that their atoms must be like sandspurs, so they stuck together while liquid atoms were more like marbles that could roll over each other. Democritus

Aristotle (384-322 BC) was a proponent of the continuum. He believed in the four elements of air, earth, water and fire. Aristotle felt that regardless of the number of times you cut a form of matter in half, you would always have a smaller piece of that matter. This view held sway for 2000 years primarily because Aristotle was the tutor of Alexander the Great. In other words, he was more

popular than Democritus. Antoine Lavoisier (1743-1794) was the first person to make good use of the balance. He was an excellent experimenter and he based his ideas on experimental evidence. He proposed the Law of Conversation of Mass which represents the beginning of modern chemistry. This Law of Conservation of Mass states that matter cannot be created or destroyed during normal chemical reactions or physical changes, it merely

rearranges or changes form. Soon after Lavoisier’s discovery, Joseph Proust (1754-1826) developed the law of definite proportions. This law stated that a chemical compound contains the same elements in exactly the same proportions by mass regardless of the size of the sample or source of the compound. In other words, a compound like salt, NaCl, is made of the same ratio of Na to Cl by mass, whether the salt is mined

from the ground, collected from the sea, or produced in the lab.

John Dalton (1776-1844) proposed the Law of Multiple Proportions. This law led directly to the proposal of the Atomic Theory in 1803. The Law of Multiple Proportions states that if two or more different compounds are composed of the same two elements, then the ratio of the masses of the second element combined with a certain mass of the first element is always a ratio of

small whole numbers.


Dalton's Atomic Theory

1) All matter is made of extremely small particles called atoms. Atoms are indivisible and indestructible.

2) All atoms of a given element are identical in mass and properties. Atoms of different elements differ in mass and properties.

3) Compounds are formed by a combination of two or more different kinds of atoms in simple-whole number ratios.

4) A chemical reaction is a rearrangement of atoms.

Modern atomic theory is, of course, a little more involved than Dalton's theory but the essence of Dalton's theory remains valid. Today we know that atoms can be destroyed via nuclear reactions but not by chemical reactions. Also, there are different kinds of atoms (differing by their masses) within an element that are known as "isotopes", but isotopes of an element have the same chemical properties.

Many heretofore unexplained chemical phenomena were quickly explained by Dalton with his theory. Dalton's theory quickly became the theoretical foundation in chemistry.

History of Atomic Structure

John Dalton thought that atoms were the smallest possible particle. We now know that the atom is made up of even smaller particles. Here are the experiments that led to the discovery of what we believe is the structure of the atom.

J. J. Thomson (1856-1940) identified the negatively charged electron in the cathode ray

tube in 1897. He deduced that the electron was a component of all matter and calculated the charge to mass ratio for the electron.

He also proposed the "plum pudding" model of the atom. In this model, the volume of the atom is composed primarily of the more massive (thus larger) positive portion (the plum pudding). The smaller electrons (actually, raisins in the plum pudding) are dispersed throughout the positive mass to maintain charge neutrality.


Robert Millikan (1868-1953) determined the unit charge of the electron in 1909 with his Oil Drop experiment at the University of Chicago. This allowed for the calculation of the mass of the

electron. In this experiment, he used charged plates to balance droplets of oil to determine the charge to mass ratio of the droplets.

Ernest Rutherford (1871-1937) proposed the nuclear atom as the result of the gold-foil experiment in 1911.

Rutherford's Gold Foil Experiment

What happened What it means

Alpha particles (positive helium nuclei) were shot at a piece of gold foil.

Most of the particles went straight through. Most of the atom is empty space.

A few particles were slightly deflected

Since alpha particles are +, they must have come close to + protons.

A very few particles were reflected, bounced back.

The alpha particles hit something dense. This is the discovery of the nucleus.


Rutherford proposed that all of the positive charge and all of the mass of the atom occupied a small volume at the center of the atom called the nucleus and that most of the volume of the atom was empty space occupied by the electrons.

Francis Aston (1877-1945) invented the mass spectrograph in 1920. He was the first person to observe isotopes. For example he observed that there were three different kinds of hydrogen atoms.

James Chadwick (1891-1974) discovered the neutron in 1932. Chadwick was a collaborator of Rutherford's. Interestingly, the discovery of the neutron led directly to the discovery of fission and ultimately to the atomic bomb. Neils Bohr (1885-1962) contributed ideas to the current atomic structure models. Bohr stated that the electrons were in energy levels in which the electrons were quantized. In other words, electrons on different energy levels have different amounts of energy.

The current atomic model is the electron cloud/probability model. The energy levels are 3 dimensional volumes of electrons around the nucleus instead of rings.


Atomic Structure Summary The atom is made of a nucleus that contains protons (+) and neutrons (neutral). Surrounding the nucleus are energy levels that hold electrons (-). The atom itself is neutral because there are equal number of protons and electrons. The nucleus is positive.

Properties of Subatomic Particles

Particles Symbols Relative charge Mass Number Location

Electron e-, 0-1e -1 0 amu Energy levels

Proton p+, 11H 1 1 amu Nucleus

Neutron n0, 10n 0 1 amu Nucleus

Note: The unit amu stands for atomic mass unit. It equals 1/12 of the carbon-12 atom. Task 2a

1. List the scientist that developed or discovered the following.

a. Discovered the nucleus b. Discovered the electron c. Oil Drop experiment d. Discovered the neutron

Counting Atoms Atoms of different elements have different numbers of protons. Atoms of the same element all have the same number of protons. The atomic number (Z) of an element is the number of

protons of each atom of that element. The atomic number is found on the periodic table. In this particular square of gold from the periodic table the top number is the atomic number. It is always a whole number. The periodic table is arranged by increasing atomic number. This means that gold has an atomic number of 79 and has 79 protons. The atomic number identifies the element. Any particle that has 79 protons is gold no matter how many electrons or neutrons it has. Remember atoms have a neutral charge so if the gold atom has

79 protons, it also has 79 electrons.

The bottom number of this particular square from a periodic table is the atomic mass or mass number. The mass number (A) is the total number of protons and neutrons that make up the nucleus of an atom. The mass number does not have to be same for every atom of an element. Atoms of the same element that have different masses are called isotopes. Since electrons affect chemical properties, different isotopes of the same element have the same chemical properties. The symbols for isotopes can be written as hyphen notation (gold-197) or nuclear notation


(19779Au). In hyphen notation the mass number is given. You have to look up the atomic number

on the periodic table. In nuclear notation, the top number is the mass number and the bottom number is the atomic number.

Notice that mass numbers on the periodic table are not whole numbers. They are weighted averages of all the isotopes of that particular element.

You can also calculate the protons, electrons and neutrons for ions. Ions are atoms that have gained or lost electrons. They have a charge. Charges are written as a superscript on the right of the symbol. 23

11Na1+

Atomic particle summary Atomic number = the number of protons in the nucleus of one atom of the element

Mass number = the number of protons and neutrons

# of protons = the atomic number

# of electrons = the # of protons for atoms, if an ion add electrons for a – charge and subtract electrons for a + charge

# of neutrons is the mass number – atomic number

Example:

Calculate the number of protons, electrons, and neutrons in the following.

3115P protons = 15 electrons = 15 neutrons = 16

Na-23 protons = 11 electrons = 11 neutrons = 12

7934Se2- protons = 34 electrons = 36 neutrons = 45

10847Ag+ protons = 47 electrons = 46 neutrons = 61


Task 2b

1. Consider the following pairs; does either pair represent a pair of isotopes?

a. 4019K and 40

18Ar

b. 9038Sr and 94

38Sr

2. Determine the number of protons, electrons, and neutrons in the following.

a. 21082Pb

b. 3416S

c. Br-80

d. 6429Cu2+

e. 157N3-

3. Write the hyphen notation and the nuclear notation for an isotope with 15 electrons and 15 neutrons.

4. Write the nuclear notation for an ion that has 53 protons, 54 electrons, and 73 neutrons.

Average Atomic Mass

Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element.

Average atomic mass = Sum of (% of each isotope)(atomic mass of each isotope)

100

Task 2c

1. Three isotopes of argon occur in nature – 3618Ar, 38

18Ar, and 4018Ar. Calculate the average

atomic mass of argon to two decimal places, given the following relative atomic mass and abundances of each of the isotopes: argon-36 (35.97 amu; 0.337%), argon-38 (37.96 amu; 0.063%), and argon-40 (39.96 amu; 99.600%).

2. Naturally occurring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotopic form of boron. What must the atomic mass of this second isotope be in order to account for the 10.81 amu average atomic mass of boron? (Write the answer to two decimal places.)


Topic 3: Electromagnetic Spectrum & Quantum Theory

Arrangement of Electrons in Atoms (Chapter 4 in Modern Chemistry)

Introduction

There were two light theories in the early 1900. Sir Isaac Newton subscribed to the particle theory of light. Christian Huygens subscribe to the wave theory of light. There was data to support both theories. Einstein developed the idea of a photon, Bohr proposed a quantized model of the atom, and eventually, Louis deBroglie, came up with the Wave-Particle Duality of Nature. He said that sometimes waves act like particles and sometimes particles act like waves. This was true of very small particles. Eventually, the Quantum theory was developed and so were electron configurations.

Newton Huygens Einstein Bohr deBroglie

Properties of Light Visible light is a kind of electromagnetic radiation, which exhibits wavelike behavior as it travels through space. There are other types of light that make up the electromagnetic spectrum. All forms of electromagnetic radiation move at a constant speed. We call this the speed of light (c); c = 3.00 x 108 m/s.


Wave motion is repetitive. It is characterized by wavelength and frequency. Wavelength (O) is the distance between corresponding points on adjacent waves. Wavelength is a length unit so it is expressed in meters, centimeters, or nanometers.

Frequency (Qis defined as the number of waves that pass a given point in a specific time, usually one second. Frequency is expressed in waves/second, s-1,or hertz (Hz).

Wavelength & frequency are inversely proportional, meaning longer wavelengths have lower frequencies and vice versa. Mathematically, they are related to each other in the following relationship.

c = OQ

The Photoelectric Effect

The photoelectric effect refers to the emission of electrons from a metal when light shines on the metal. It was found that only light above certain minimum frequencies could cause these electrons to be released no matter how intense the light. This could not be explained by the wave theory.

Max Planck suggested that energy is emitted or absorbed not as continuous waves, but as small, specific packets of energy called quanta. A quantum of energy is the minimum quantity of energy that can be lost or gained by an atom.

E = hQE is the energy, in joules Q is the frequency in s-1 or Hz Max Planck h is Planck’s constant;; h = 6.63 x 10-34 Js or J/Hz

By combining the equations, c = OQand E = hQ, another useful equation can be derived.

E = hc O


Task 3a

1. Characterize each of the following as absorption or emission: a. an electron moves from E2 to E1 b. an electron moves from E1 to E3 c. an electron moves from E6 to E3

2. Which energy-level change above emits or absorbs the highest energy? The lowest

energy?

3. Solve the following problems using the equations in this section. a. Determine the frequency of light whose wavelength is 4.257 x 10-7 cm.

b. Determining the energy in joules of a photon whose frequency is 3.55 x 1017 Hz.

c. When sodium is heated; a yellow spectral line whose energy is 3.37 x 10-19 J per photon is produced. What is the wavelength of this light?

d. The laser in an audio compact disc player uses light with a wavelength of 7.80 x 102 nm. Calculate the frequency of this light. Calculate the energy of a single photon of this light.

Rydberg Equation

You can also determine the amount of energy absorbed or emitted as electrons move from one energy level to another using a form of the Rydberg equation.

UE = Ef – Ei = Ephoton

E = the energy associated with a particular quantum number, n. By calculating the energies associated with two different quantum levels and finding the difference, one can calculate the energy required to promote an electron from one level to another, or calculate the energy released when an electron falls back from a higher level to a lower level. Energy emitted is exothermic and is given a negative charge. Energy absorbed is endothermic and is given a positive charge.

Task 3b

1. What is the energy difference when an electron moves from E3 to E5? Is the energy emitted or absorbed. Use the appropriate sign.

E = - 2.178 x 10-18

n2

U E = ( -2.18 x 10-18 J ) ( 1 - 1 )

nf2 ni

2


The Hydrogen-Atom Line-Emission Spectrum

When electricity is passes through a gas at low pressure, the energy of some of the gas atoms increases. The lowest energy state of an atom is its ground state. A state in which an atom has a higher energy than it has in its ground state is an excited state. For an electron to move from its ground state to an excited state, energy must be absorbed. When an electron falls from an excited state to a lower excited state, or to its ground state, energy will be released. This energy is released in the form of light called a photon. The production of colored light in neon lights is an example of this process.

When electricity was passed through hydrogen gas at low pressure, a pinkish glow was given off. When a narrow beam of the emitted light was shined through a prism, it was separated into four specific colors of the visible spectrum. The four bands (lines) of light were part of what is known as hydrogen’s line-emission spectrum.

These distinct bands at specific wavelength & frequencies indicated that the energy differences between the atoms’ energy states were fixed. This led to the Bohr model of the atom. In this model, Niels Bohr linked the atom’s electron to photon emission. The electron can circle the nucleus only in allowed paths, or orbits. When the electron is in one of these orbits, the atom has a definite, fixed energy.

Here are two animations that will help you visualize what happens to electrons as they absorb or emit energy.

http://science.sbcc.edu/physics/solar/sciencesegment/bohratom.swf http://physics.gac.edu/~chuck/PRENHALL/Chapter%2031/AABXTEJ0.html?I1.x=40&I1.y=17

The Quantum Model of the Atom

So, after the photoelectric effect and hydrogen’s line-emission spectrum revealed that light could behave as particles and waves, Louis de Broglie proposed the wave-particle duality of nature. Then Werner Heisenberg tried to determine where the electrons were in the atom. Because you need a photon to


detect an electron, and a photon causes an electron to be knocked off course, there is always an uncertainty about the location of an electron. This is called the Heisenberg uncertainly principle. It states that it is impossible to determine simultaneously both the position and velocity of an electron or any other particle. There are only probable positions of electrons.

Max Planck founded the Quantum theory, which describes mathematically the wave properties of electron and other very small particles. This theory states that electrons do not travel around the nucleus in neat orbits, as Bohr postulated, but exist in certain regions or volumes, called orbitals. An orbital is a three-dimensional region around the nucleus that indicates the probable location of an electron.

s orbital

Atomic Orbitals and Quantum Numbers

In order to completely describe orbitals, scientists use quantum numbers. Quantum numbers specify the properties of atomic orbitals and the properties of electron in those orbitals. There are four quantum numbers. The first three indicate the main energy level, the shape, and the orientation of an orbital. The fourth, describes the spin on an electron in the orbital.

Quantum number notes

There are 4 Quantum Numbers: n, A, m, s

1. n = Principal Quantum Number (n = 1, 2, 3…)

x Represents the main energy level, if n = 1, 1st energy level, n = 2, 2nd energy level, etc.

x This also represents the size of the electron cloud. x n = 1 is the energy level closest to the nucleus, the larger n, the farther away from

the nucleus x 2n2 – determines maximum number of electrons that can occupy an energy level

2. A= sublevel (A = 0 to n-1)

x Refers to different energy states in each energy level x Determines the shape of the orbital(s) x Number of sublevels = n

o n = 1, A has 1 sublevel Æ s (A = 0)

o n = 2, A has 2 sublevels Æ s (A = 0), p (A = 1)


o n = 3, A has 3 sublevels Æ s (A = 0), p (A = 1), d (A = 2)

o n = 4,A has 4 sublevels Æ s (A = 0), p (A = 1), d (A = 2), f (A = 3)

3. m = orbital (m = -A to +A)

x Space occupied by a pair of electrons x Determines orientation in space x Each orbital can be represented by a box x Each orbital can hold a maximum of two electrons x s = 1 orbital, 1 box, a maximum of 2 electrons (0)

x p = 3 orbitals, 3 boxes, a maximum of 6 electrons (-1,0,1)

x d = 5 orbitals, 5 boxes, a maximum of 10 electrons (-2, -1, 0, 1, 2)

x x x

x f = 7 orbitals, 7 boxes, a maximum of 14 electrons (-3, -2, -1, 0, 1, 2, 3)


4. s = spin x in order for 2 electrons to occupy the same orbital, they must have opposite spins x the box represents the orbital, the arrows represent electrons spinning in opposite

directions, there cannot be arrows pointing in the same direction in the same box Each arrow represents an electron (+ ½, -½)

Task 3c

Answer the following questions based on quantum numbers.

1. How many quantum numbers are there? 2. What is the maximum number of electrons in an orbital? 3. What is the maximum number of electrons in n = 2? 4. Which quantum number represents the shape of the electron cloud? 5. Which quantum number represents the volume of the electron cloud? 6. How many orbitals are in the p sublevel? 7. How many electrons can be in the d sublevel? 8. Which energy level has the lowest energy? 9. Which sublevel has the lowest energy? 10. How many sublevels are in n = 4? What are they?

Rules for placing electrons in orbitals

Pauli’s Exclusion Principle

x No two electrons in an element can have the same set of quantum numbers.

Aufbau Principle

x Electrons occupy orbitals of lowest energy first – 1s

Hund’s Rule

x Within a sublevel, orbitals are half-filled with electrons before they become filled x p sublevel with 3 electrons


n A m s Honors (ex.)

1 s 1

1,0,0,+ ½ 1,0,0,- ½

2 s, p 1, 3

2,0,0,+ ½ 2,0,0,- ½ 2,1,-1, + ½ 2,1,-1, - ½ 2,1,0, + ½ 2,1,0, - ½ 2,1,1, + ½ 2,1,1, - ½

3 s, p, d 1, 3, 5

4 s, p, d, f 1, 3, 5, 7

Look up the atomic number of the element for which you are drawing the orbital filling diagram. Since these are atoms, this not only tells you the number of protons but also the number of electrons. The electrons are represented by arrows. For example, K has atomic number of 19, so its atoms have 19 electrons, therefore, 19 arrows. For example the orbital filling diagrams for for H is: For He: 1s1 1s2 For Li: For Be: 1s2 2s1 1s2 2s2


For B: For N:

1s2 2s2 2p1 1s2 2s2 2p3 For F:

1s2 2s2 2p5 For Si:

1s2 2s2 2p6 3s2 3p2 Note that you always start with 1s. Always start with an “up” arrow. When you are starting to fill up the orbitals (boxes) you placed “up” arrows in all the boxes until there is at least one arrow in the orbital, then you go back and finish filling that orbital. Only two arrows can go in a box. You can also write the electron configurations for the same elements. You just write the numbers below without drawing the boxes. For H: 1s1 (read as: one s one) For He: 1s2 For Li: 1s2 2s2 For Be: 1s2 2s2 For B: 1s2 2s2 2p1 For N: 1s2 2s2 2p3 For F: 1s2 2s2 2p5 For Si: 1s2 2s2 2p6 3s2 3p2

It is very important that you write the electron configuration in order of lowest energy to highest energy. You can use the diagonal rule or the periodic table.

The diagonal rule shows the order of electron filling. Follow the arrows.


Task 3d

1. Draw the orbital filling diagram for the following.

a. P

b. V

c. Na

2. Write the electron configuration for the following.

a. W

b. Zn

c. Ca

d. Tl

e. Br

3. What are the quantum numbers for the last electron in each of the following?

a. As

b. Nb

c. Ba

4. Four electrons in an atom have the four sets of quantum numbers given below. Which electrons are in the same orbital? Explain your answer.

a. 1, 0,0, -___

b. 1, 0, 0, +___

c. 2, 1, 1, +___

d. 2, 1, 0, +___

5. Which of the sets of quantum; numbers below are possible? Which are impossible? Explain your choices.

a. 2, 2, 1, +___

b. 2, 0, 0, -___

c. 2, 0, 1, -___

You also need to be able to write the quantum numbers for electrons. In the notes above there were numbers in parentheses. These are the quantum numbers. For example: Li has an electron configuration of the quantum numbers for the last electron would be 2, 0, 0, +½. 1s2 2s1


The 2 represents the 2 level, the first 0 represents the s sublevel, the second 0 represents the box, and the + ½ represents the clockwise spin. (We actually do not know if it is clockwise or counterclockwise, by convention we use + ½ for the up arrow.) Another example: Nitrogen’s orbital filling diagram is 1s2 2s2 2p3 The quantum numbers for the last electron would be 2, 1, -1, +½.

Electron Configuration & the Periodic Table (Chapter 5 in Modern Chemistry)

You can also write electron configurations using the periodic table. Below is a periodic table labeled with the appropriate blocks. This is the method I use, and the method taught in most college classes.

1 2 3 4 5 6 7

The rows on the periodic table represent the outer energy level or principle quantum number. The first two columns are the s blocks plus He. The middle section (transition metals) is the d block elements. Notice that the principle quantum number decreases by one in this section. In other words a d block element on row 5, has a 4d sublevel. The last 6 columns of the periodic table are the p block elements. Notice that the principle quantum number for these elements is the same as the row number. The f section of the periodic table is located on the bottom two rows of the table.


When using the periodic table to write electron configurations, follow the chart, counting the elements in each block until you reach the element you are writing the configuration for. Alternately, you can write the noble gas configuration, by writing the noble gas immediately before the specified element in brackets [ ]. Here are some examples using the periodic table above. Electron Configuration Noble gas configuration For Sr: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 [Kr] 5s2 For Al: 1s2 2s2 2p6 3s2 3p1 [Ne] 3s2 3p1 For Pb: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p2 [Xe] 6s2 4f14 5d10 6p2

Now you try some.

Task 3e

1. Write the quantum numbers for the last filling electron in the following.

a. K

b. Co

c. Sn

2. Write the electron configuration for the following using the periodic table.

a. O

b. Mo

c. Cs

3. Write the noble gas configuration for the following.

a. Mg

b. Fe

c. Cl

d. Au

e. Fr


4. Write the symbol of the element that is represented below by the electron configuration or noble gas configuration.

a. 1s2 2s2 2p6 3s2 3p6 4s2 3d8

b. [Kr] 5s2 4d10 5p3

c. 1s2 2s2 2p6 3s2 3p4

Exceptions to the electron configuration

There are some elements that do not follow the general rules for writing electron configuration, for example, Cr and Cu. The anomalies of Cr and Cu are easy to explain once you know that a half-filled or completely filled d shell is considered to have extra stability. Hence configurations ending 4s1 3d5 and 4s1 3d10 rather than 4s2 3d4 and 4s2 3d9 are considered to be preferable. In each case one of the s electrons is promoted to the d shell to create a more stable configuration.

Valence Electrons

Valence electrons are the electrons that are located on the outside energy level. You can tell which electrons are valence by the main energy level. For example, in the electron configuration, 1s2 2s2 2p6 3s2 3p4, the highest energy level is 3. There are 6 electrons in level three, two in the s sublevel, and 4 in the p sublevel.

Electron Dot Diagrams

Electron Dot Diagrams are visual representations of the valence electrons in an atom. For example: Al has a noble gas configuration of [Ne] 3s2 3p1. If I were to draw the orbitals for the valence electrons only they would be .

3s 3p

There are two electrons in the 3s and 1 electron in the first orbital of the 3p. I could draw the valence electrons around the symbol.

Al

The electron dot diagram for Al would be : Al . It doesn’t matter where you put the electrons only that you put the correct number of electrons in each box. It is called electron dot because you use dots instead of arrows. In the electron dot diagram, the symbol represent the nucleus and all the inner electrons. The dots represent the valence electrons.


Here’s another example: Te [Kr] 5s2 4d10 5p4 (Remember 4d is not valence) 5s 5p The electron dot would be

Task 3f

1. Using the periodic table, how many valence electrons are in the following elements?

a. Br

b. Sr

c. Ar

d. Fr

2. Draw the electron dot diagram for each atom in (1).

Electron Configurations of Ions

You can also write the electron configurations of ions. The positive ions have lost electrons. Electrons can only be removed from the outside energy level. For example, Na has an electron configuration of [Ne] 3s1. Na+ has an electron configuration of [Ne] or preferably [He] 2s2 2p6. Fe has an electron configuration of [Ar] 4s2 3d6. Fe2+ has an electron configuration of [Ar] 3d6. Fe3+ has an electron configuration of [Ar] 3d5. Negative ions have gained electrons. Electrons should be placed in the next available orbital. For example: S has and electron configuration of [Ne] 3s2 3p5. Cl- has an electron configuration of [Ne] 3s2 3p6.

Isoelectronic Species

Species that have the same electron configuration are called isoelectronic. These usually come in a series. For example: Ne is isoelectronic with O2-, F-, Na+, and Mg2+.

Predicting Oxidation Numbers

An oxidation number is the tendency of an atom to gain or lose electrons. According to the octet rule atoms are more stable whenever they have 8 valence electrons (except for things in period 1). To determine the oxidation number, you need to know how many electrons are gained or lost, or how many are likely to be gained or lost.


For example: Ca has a configuration of [Ar] 4s2. It is easier to lose those 2 valence electrons than to gain 6 to make 8. Losing electrons cause the charge to be positive, so the oxidation number of Ca that is most likely is 2+. The calcium ion is usually written as Ca2+.

P has a configuration of [Ne] 3s2 3p3. It has 5 valence electrons. It is easier to gain three electrons to make a total of eight than to lose the 5 it has. So it will gain 3 electrons, making it a negative three charge. This is written as P3-.

Task 3g

1. Without looking at the periodic table, identify the group, period, and block in which the element that has the electron configuration [Xe] 6s2 is located.

2. Without looking at the periodic table, write the electron configuration for the Group 1 element in the third period. Is this element likely to be more reactive or less reactive than the element described in (1)?

3. Write the electron configurations for the following ions.

a. Cl-

b. K+

c. Fe2+

d. Fe3+

e. P3-

4. Which of the ions in (3) are isoelectronic with each other?

5. Which of the following does not have the same configuration as a noble gas: Na+, Rb+, O2-, Br-, Ca+, Al3+, S2-?

6. What is the probable oxidation number for the following elements? a. Mg b. K c. S d. I e. P

7. What is the electron configuration of silver? It only has one oxidation number, what is it?

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Topic 4: Periodic Table & Trends The Periodic Law (Chapter 5 in Modern Chemistry)

The History of the Periodic Table

Stanislao Cannizzaro discovered a method of accurately measuring atomic masses.

Dmitri Mendeleev is credited with organizing the first periodic table based on atomic masses.(1869) He noticed the when the elements were arranged in order of increasing atomic mass, certain similarities in their chemical properties appeared at regular intervals. These repeating patterns are referred to as periodic. Mendeleev did find some discrepancies, he placed iodine after tellurium even though based on atomic masses they should be reversed.

Tellurium acted more like O, S, and Se. Iodine acted more like F, Cl, and Br. He knew there was some problems with using atomic masses for ordering the table. He also left blanks in his periodic table. He boldly predicted the existence and properties of the elements that would fill three of these blanks based on the properties of elements that were similar in his table. Eventually, all three of these elements were discovered, Sc, Ga, and Ge. Their properties were strikingly similar to those predicted by Mendeleev.

Henri Mosely discovered that the elements in the periodic table fit into patterns better when they were arranged in increasing order according to nuclear charge, or number of protons (atomic number). (1911) This corrected the discrepancies in Mendeleev’s table.

Periodic Law state the physical and chemical properties of the elements are periodic functins of their atomic numbers.

The Modern Periodic Table is an arrangement of the elements in order of their atomic numbers so that elements with similar properties fall in the same column, or group. Remember that the numbers across the top represent the group, while the numbers down the side represent the period.

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The Element Song by Tom Lehrer http://www.privatehand.com/flash/elements.html

Noble gases (Group 18)

All atoms of the noble gases have their outer s and p orbitals filled. We will see later that these atoms require very large amounts of energy to form ions, so much in fact, that they are difficult to alter chemically and as such are inert (unreactive) and do not tend form ions.

Alkali metals (Group 1) & Alkaline Earth metals (Group 2) METALS

Group 1 atoms have an electronic structure [Noble gas] ns1. This means that they tend to lose the s electron when they from an ion, leaving behind an inert noble gas type structure. This explains why Group 1 elements tend to only form 1+ ions. Group 2 atoms have an electronic structure [Noble gas] ns2. This means that they tend to lose the two s electrons when they from an ion, leaving behind an inert noble gas type structure. This explains why Group 2 elements tend to only form 2+ ions.

A similar argument can be applied to group 3 atoms and their simple ions.

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Groups 16 & 17 (Chalcogens & Halogens) NON-METALS

Group 16 atoms have an electronic structure [Noble gas] ns2 np4. This means that they tend to gain two p electrons when they from an ion, to reach an inert noble gas type structure with a charge of 2-. Check out this animation: http://www.privatehand.com/flash/oxygen.html . It is about oxygen. Very cute! Group 17 atoms have an electronic structure [Noble gas] ns2 np5. This means that they tend to gain one p electron when they from an ion, to reach an inert noble gas type structure with a charge of 1-.

Task 4a

1. What name is given to each of the following groups of elements in the periodic table? a. Group 1 b. Group 2 c. Groups 3-12 d. Group 17 e. Group 18

2. Based on what you know about their electron configurations, which groups do you think

are the most active? Why? The least active? Why?

Periodic Properties As you have learned earlier, the elements in the same group have similar electron configurations, therefore they also have similar physical and chemical properties because their valence electrons are the same. You need to be able to identify the properties and how they relate to each other based on their placement on the periodic table.

Before we discuss the periodic property trends, we need to discuss two reasons that properties change based on the periodic table.

The shielding effect causes properties to change that are in the same group (column). The shielding effect occurs because the inner electrons shield the nucleus from the outer electrons. The more inner electrons there are between the valence electrons and the nucleus then the smaller the attraction of the nucleus on the outer electrons. Even though the valence electrons are being attracted to the nucleus, they are also being repelled by all the inner electrons.

The effective nuclear charge causes properties to change that are in the same period (row),. This means that the positive charge of the nucleus is increasing while the number of energy levels is the same. Since the electrons are negatively charged, they are attracted to nuclei, which

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are positively charged. Many of the properties of atoms depend not only on their electron configurations but also on how strongly their outer electrons are attracted to the nucleus. The distance from the nucleus is not changing but the charge of the nucleus is.

Atomic radii

Atomic radii refers to the size of an atom. It is actually defined as one-half the distance between the nuclei of identical atoms that are bonded together. Atomic radii increases down a group and decreases across a period. It increases toward Fr. The size of the atom naturally increases down a group because the volume of the electron cloud gets bigger as the number of energy levels increase. The size of the atom gets smaller as the atomic number increases because the charge of the positive nucleus is getting larger and attracts the electrons more and more. It is true that the number of electrons is also increasing, but they are all placed in the same energy level, meaning that they are the same distance from the nucleus. The higher the nuclear charge the stronger the

pull on those electrons.

Task 4b

1. Referring to the periodic table, arrange the following atoms in order of increasing size: P, S, As, Se.

2. Of cesium, Cs, hafnium, Hf, and gold, Au, which element has the smallest atomic radius? Explain your answer in terms of trends in the periodic table.

Ionic Radii

A positive ion is known as a cation. The formation of a cation by the loss of one or more electrons always leads to a decrease in atomic radius because as electrons are removed the nucleus has a greater attraction for the electrons that remain. The positive ion has lost a whole layer of electrons.

A negative ion is known as an anion. The formation of an anion by the addition of one or more electrons always leads to an increase in atomic radius. The extra electrons cause the electron

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cloud to spread out due to greater electron repulsion while the attraction from the protons remains the same.

Positive ions are smaller than the atom from which they came, while negative ions are larger than the atom from which they came.

Task 4c

1. Distinguish between a cation and an anion.

2. Which of the following cations is least likely to form: Sr2+, Al3+, K2+?

3. Which of the following anions is least likely to form: I-, Cl2-, O2-?

4. From each set, determine which atom or ion is the largest.

a. Ca, Ca2+

b. Cl, Cl-

c. S, S-, S2-

d. Sr, Sr+, Sr2+

e. K+, Ca2+

f. N3-, O2-, F-

g. Ca2+, Sr2+, Ba2+

h. O2-, S2-, Se2-

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Ionization Energy

An electron can be removed from an atom if enough energy is supplied. Any process that results in the formation of an ion is referred to as ionization. Suppose A is any element.

A + energy Æ A+ + e-

A+ + energy Æ A2+ + e-

Ionization Energy is the amount of energy required to remove one electron from a neutral atom of an element to make an ion. Specifically, this is the first ionization energy or IE1. Ionization energy increases toward F.

Ionization energy decreases down a group because the farther the valence electron is from the nucleus the lower the attraction between them. Remember that as you move down a group the period number increases, meaning there is another level of electrons being added. Therefore, there will be more shielding electrons between the nucleus and the valence electrons causing repulsion. So, the farther down the element is in the group the lower the ionization energy.

Ionization energy increases across a period because the effective nuclear charge (the positive charge) is getting larger while the shielding effect (levels of electrons) remains the same. This means there will be a stronger attraction between the valence electrons and the nucleus, so it will take more energy to remove an electron.

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The second ionization energy is the amount of energy required to remove the second electron from an ion. The third ionization energy is the amount of energy required to remove the third electron from an ion, etc. Successive ionizations require more energy. The more electrons that are removed the harder it is to remove the next. Each successive electron removed from an ion feels an increasingly stronger effective nuclear charge.

Notice that there is a big jump in the IE1 to IE2 of sodium. Sodium has only one valence electron. It is relatively easy for it to lose that electron when bonding. It will then have a stable configuration with 8 valence electrons. It will be much more difficult to remove the second electron from sodium because this will make the ion become unstable so it takes a lot more energy to do this. Basically, it is easier to remove valence electrons than it is to remove the electron immediately after the last valence electron.

Task 4d

1. Referring to the periodic table, arrange the following atoms in order of increasing first ionization energy: Ne, Na, P, Ar, K.

2. In general ionization energy increases toward F. Refer to the graph on ionization energy trends. Considering electron configurations, why do you think B has a lower IE1 than Be? O has a lower IE1 than N?

3. Write the equations that show the process for the following.

a. The first ionization energy for tin

b. The second ionization energy for the tin(I) ion

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4. Explain each of the following.

a. Why does Li have a larger first ionization energy than Na?

b. The difference between the third and fourth ionization energies of scandium is much larger than the difference between the third and fourth ionization energies of titanium. Why?

c. Why does Li have a much larger second ionization energy than Be?

5. Here are the ionization energies for an element in period 2: 900, 1757, 14849, 21007. Which element is represented by these energies?

Electron Affinity

Neutral atoms can also acquire electrons. The energy change that occurs when an electron is acquired by a neutral atom is called the atom’s electron affinity. Most atoms release energy when they acquire an electron. Since energy is given off it will have a negative sign (exothermic).

A + e- Æ A- + energy

Electron affinity increases toward F. It decreases down a group because there is an increase in atomic radius down a group, which decreases electron affinities. Therefore the nucleus is not as likely to gain an electron.

Noble gases have a 0 electron affinities. They have a full valence level; therefore they do not gain electrons.

Across the periods, the effective nuclear charge is increasing. The shielding effect remains essentially the same. The larger nuclei tend to be more likely to attract electrons other than its own than nuclei that are smaller.

Task 4e

1. Order the atoms in each of the following sets from the least electron affinity to the most.

a. O, S

b. F, Cl, Br, I

c. N, O, F

d. B, C, N

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Electronegativity

Electronegativity is a measure of the ability of an atom in a chemical compound to attract electrons from another atom in the compound. Fluorine is the most electronegative element. It is arbitrarily assigned a value of four. The other values are calculated in relation to this value.

Electronegativities increase toward F.

Note that the noble gases have a 0 electronegativity. Remember that they are very stable and do not normally form compounds so they do not pull additional electrons toward them.

The alkali and alkaline-earth metal are the least electronegative. In compounds, their atoms have a low attraction for electrons.

Nitrogen, oxygen, and the halogens are the most electronegative elements. Their atoms attract electrons strongly in compounds.

Electronegativities tend to either decrease down a group or stay the same.

Task 4f

1. Using the periodic table, place the following element in order of increasing electronegativity: O, Fe, Ge, Sr, S, Zr

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Other Periodic Properties

Valence electrons

The number of valence electrons can be determined by the group on the periodic table. Notice that the number of valence electrons is related to the group number.

Group # of valence

electrons

1 1 2-12 2 (vary) 13 3 14 4 15 5 16 6 17 7 18 8 (except He, 2)

Oxidation Numbers

Oxidation numbers also vary in a regular pattern on the periodic table. Notice that groups 1-13 tend to lose electrons to become positive. Group 14 can either lose or gain 4 electrons. Groups 15-17 gain electrons to become negative. Group 18 is stable, so it neither loses nor gains electrons.

Group # of valence

electrons

1 1+ 2-12 2+ (vary with RN) 13 3+ 14 4+,4- 15 3- 16 2- 17 1- 18 0

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Topic 5: The Language of Chemistry Chemical Formulas & Chemical Compounds

(Chapter 7 in Modern Chemistry)

A Chemical Formula

Recall that a chemical formula indicates the relative number of atoms of each kind in a chemical compound. For a molecular compound, the chemical formula reveals the number of atoms of each element contained in a single molecule of the compound.

C8H18

Unlike a molecular compound, an ionic compound consists of a lattice of positive and negative ions held together by mutual attraction. The chemical formula for an ionic compound represents one formula unit – the simplest ratio of the compound’s positive ions (cations) and its negative ion (anion).

Al2(SO4)3

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Note how the parentheses are used. They surround the polyatomic anion to identify it as a unit. When there is no subscript written next to an atoms’ symbol, the value of the subscript is understood to be one.

Monatomic Ions

By gaining or losing electrons, many main-group elements form ions with noble-gas configurations (recall from Topics 3 & 4). Ions formed from a single atom are known as monatomic ions.

Monatomic cations are identified simply by the element’s name. Naming monatomic anions is slightly more complicated. First, the ending of the element’s name is dropped. Then the ending –ide is added to the root name.

Examples of Cations

Examples of Anions

Element Cation

Element Cation K K+

F F-

Potassium Potassium ion

Fluorine Fluoride ion

Mg Mg2+

N N3- Magnesium Magnesium ion

Nitrogen Nitride ion

The oxidation numbers of main-group monatomic ions can be determined by looking at their group number on the periodic table. The names of many of the ions include Roman numerals. These numerals are part of the Stock system of naming chemical ion and elements. They are used for elements that have more than one oxidation number.

Task 5d

1. Label the following as a cation or an anion.

a. Te

b. W

c. Fr

2. Name the following ions.

a. Ca2+ d. N2+

b. Cu+ e. P3-

c. Se2- f. I-

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Some Common Monatomic Ions

Main-group elements

1+ 2+ 3+ lithium Li+ beryllium Be2+ aluminum Al3+ sodium Na+ magnesium Mg2+ potassium K+ calcium Ca2+ rubidium Rb+ strontium Sr2+ cesium Cs+ barium Ba2+

1- 2- 3- fluoride F- oxide O2- nitride N3- chloride Cl- sulfide S2- phosphide P3- bromide Br- iodide I-

d-Block elements and others with multiple ions

1+ 2+ 3+ 4+

copper(I) Cu+ vanadium(II) V2+ vanadium(III) V3+ vanadium(IV) V4+

silver Ag+ chromium(II) Cr2+ chromium(III) Cr3+ tin(IV) Sn4+

manganese(II) Mn2+ iron(III) Fe3+ lead(IV) Pb4+

iron(II) Fe2+ cobalt(III) Co3+

cobalt(II) Co2+

nickel(II) Ni2+

copper(II) Cu2+

zinc Zn2+

cadmium Cd2+

tin(II) Sn2+

lead(II) Pb2+

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Binary Ionic Compounds

Binary compounds are made of two elements, a metal and a non-metal. In a binary ionic compound, the total numbers of positive charges and negative charges must be equal. In order to write the formula of an ionic compound, you need to write the cation and the anion, then balance the charges so that the sum of the charges equal zero.

For example:

Mg and Cl: Mg2+, Cl- There must be 2 Cl’s in order for the negative charge to equal 2-. 2+ and 2- equal zero. The correct formula for Mg and Cl is MgCl2.

Al and O: Al3+, O2- These numbers will not go into each other so I find a number that both of them will go into, 6. I will need 2 Al’s to have a charge of 6+ (2 x 3+ = 6+). I will need 3 O’s to have a charge of 6- (3 x 2- = 6-). The correct formula for Al and O is Al2O3.

Naming Binary Ionic Compounds

The nomenclature, or naming system, of binary ionic compounds involves combining the names of the compound’s positive and negative ions. To name binary compounds, follow the rules below.

1. Name the cation. (First ion, metal)

2. Name the anion. (Second ion, non-metal) Remember to change the ending of the last ion to –ide. For example: KBr is potassium bromide MgCl2 is magnesium chloride

K2O is potassium oxide Task 5e

1. Write the formulas for the binary compounds formed between the following elements;

a. potassium and iodine d. aluminum and sulfur b. sodium and sulfur e. aluminum and nitrogen c. lithium and phosphorus f. barium and oxygen

2. Name the binary compounds indicated by the following formulas:

a. AgCl d. SrF2

b. ZnO e. CaO

c. CaBr2 f. Ba2P3

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The Stock System of Nomenclature

Some elements, such as iron, form two or more cations with different charges. To distinguish the ions formed by such elements, scientists use the Stock system of nomenclature. This system uses a Roman numeral to indicate an ion’s charge. The numeral is enclosed in parentheses and placed immediately after the metal name.

Fe2+ Fe3+

iron(II) iron(III)

Names of metals that commonly form only one cation do not include a Roman numeral.

Na+ Ba2+ Al3+

sodium barium aluminum

There is no element that commonly forms more than one monatomic anion.

Naming a binary ionic compound according to the Stock system is illustrated below:

CuCl2

copper(II) chloride

If you are writing the name you must figure the charge to put with the cation. It is best to start at the back of the compound with the anion, to determine what the oxidation number of the cation will be.

Task 5f

1. Write the formula and give the name for the compounds formed between the following ions:

a. Cu2+ and Br- d. Hg2+ and S2-

b. Fe2+ and O2- e. Sn2+ and F-

c. Pb2+ and Cl- f. Fe3+ and O2-

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2. Give the names for the following compounds:

a. CuO c. SnI4

b. CoF3 d. FeS

Compounds Containing Polyatomic Ions

First, you must learn your polyatomic ions. Do Not memorize. You will need these all year.

Some Common Polyatomic Ions

1+ *ammonium NH4

+

1- 2- 3-

* acetate CH3COO- (C2H3O2

-) *carbonate CO32- *phosphate PO4

3-

*bromate BrO3- *chromate CrO4

2- *chlorate ClO3

- * dichromate Cr2O72-

* chlorite ClO2- *sulfate SO4

2-

*cyanide CN- *sulfite SO32-

*hydroxide OH-

*iodate IO3-

*nitrate NO3-

* nitrite NO2-

*permanganate MnO4-

Notice that most of the polyatomic ions are negatively charged and most are oxyanions – polyatomic ions that contain oxygen. Some elements can combine with oxygen to form more than one type of oxyanion. Learn all the oxyanions that end in –ate and you’ll be able to figure the other oxyanions of that element. For example:

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Ternary compounds are made up of more an element and a polyatomic ion.

To name ternary compounds, you must know your polyatomic ions and follow the rules below.

1. Name the cation.

2. Name the anion. If the anion is an element from the periodic table, change the ending to –ide. If the anion is a polyatomic ion, do not change the ending.

For example: NH4Cl is ammonium chloride CaCO3 is calcium carbonate K2SO4 is potassium sulfate

Writing formulas of ternary compounds

Just as in binary ionic compounds, the sum of the positive and negative ions must equal zero. When multiples of a polyatomic ion are present in a compound, the formula for the polyatomic ion is enclosed in parentheses as in aluminum sulfate, Al2(SO4)3.

Task 5g

1. Write the formula for the following ternary compounds.

a. Lithium nitrate

b. Copper(II) sulfate

c. Sodium carbonate

d. Calcium nitrite

e. Potassium perchlorate

2. Give the names for the following compounds.

a. Ca(OH)2

b. KClO3

ClO4- perchlorate 1 oxygen more than the -ate ion

ClO3- chlorate the -ate ion

ClO2- chlorite 1 oxygen less than the -ate ion

ClO- hypochlorite 2 oxygens less than the -ate ion

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c. NH4OH

d. Fe2(CrO4)3

e. KClO

Molecular Compounds

Molecular binary compounds are made of two non-metals. Unlike ionic compound, molecular compounds are composed of individual covalently bonded units, or molecules. Chemists use two nomenclature systems to name binary molecules.

Naming Molecular Compounds with Prefixes

The old system of naming molecular compounds is based on the use of prefixes. For example, the molecular compound CCl4 is named carbon tetrachloride. The prefix tetra- indicates that four chloride atoms are present in a single molecule of the compound.

The rules for the prefix system of nomenclature of binary molecular compounds are as follows.

1. The element that has the smaller group number is usually given first. If both elements are in the same group, the element whose period number is greater is given first. This element is given a prefix only if it contributes more than one atom to a molecule of the compound.

2. The second element is named by combining (a) a prefix indicating the number of atoms contributed by the element, (b) the root of the name of the element, and (c) the ending -ide.

3. The o or a at the end of a prefix is usually dropped when the word following the prefix begins with another vowel.

P4O10

Tetraphosphorus decoxide

Numerical Prefixes

Number Prefix 1 mono- 2 di- 3 tri- 4 tetra- 5 penta- 6 hexa- 7 hepta- 8 octa- 9 nona-

10 deca-

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Task 5h

1. Name the following binary molecular compounds using prefixes.

a. CI4 e. As2S3

b. SO3 f. NCl3

c. PCl3 g. SO2

d. PCl5 h. ClBr

2. Write the formulas for the following compounds:

a. Carbon tetrachloride c. oxygen difluoride

b. Dinitrogen trisulfide d. sulfur hexafluoride

Naming Molecular Compounds with the Stock System

In order to name molecular compounds using the stock system, or for that matter even ionic compounds using the stock system, you have to be able to assign oxidation numbers to the elements in the compound. Namely, you must be able to assign the oxidation number to the first element in the compound. These oxidation numbers, indicate the general distribution of electron among the bonded atoms in a molecular compound or a polyatomic ion. Oxidation numbers are also called oxidation states.

A list of rules for assigning oxidation numbers follows.

Binary Compounds of Nitrogen and Oxygen

Formula Prefix-system name

N2O dinitrogen monoxide

NO nitrogen monoxide

NO2 nitrogen dioxide

N2O3 dinitrogen trioxide

N2O4 dinitrogen tetroxide

N2O5 dinitrogen pentoxide


Rules for Assigning Oxidation Numbers

As a general rule in assigning oxidation numbers, shared electrons are assumed to belong to the more electronegative atom in each bond.

1. The atoms in a pure element have an oxidation number of zero. 2. The more electronegative element in a binary molecular compound is assigned the

number equal to the negative charge it would have as an anion. 3. Fluorine has an oxidation number of -1 in all of its compounds because it is the most

electronegative element. 4. Oxygen has an oxidation number of -2 in almost all compounds. Exceptions include

when it is in peroxides, such as H2O2, in which its oxidation number is -1, and when it is in compounds with fluorine.

5. Hydrogen has an oxidation number of +1 in all compounds containing elements that are more electronegative than it; it has an oxidation number of -1 in compounds with metals.

6. The algebraic sum of the oxidation numbers of all atoms in a neutral compound is equal to zero.

7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

Here are some examples.

UF6 Start with the most electronegative (last), F is -1 (rule 3).

Multiply that oxidation number by the number of fluorine atoms. 6 x -1 = -6.

Sum must equal zero (rule 6), so U must be +6

H2SO4 Oxygen must have a -2 oxidation number (rule 4). Multiply -2 x 4 = -8

Hydrogen must have a +1 oxidation number (rule 5). +1 x 2 = +2

The sum must equal zero (rule 6), so S must equal +6.

ClO3- Oxygen must have a -2 oxidation number (rule 4). Multiply -2 x 3 = -6.

The sum of a polyatomic ion equals its charge, in this case, -1. That means the oxidation number of chlorine is +5.


Task 5 i

1. Assign oxidation numbers to each atom in the following compounds or ions.

a. HF e. CS2

b. CI4 f. H2CO3

c. H2O g. NO2-

d. PI3 h. SO42-

2. Name the following binary molecular compounds using the Stock system.

a. CI4 e. As2S3

b. SO3 f. NCl3

c. PCl3 g. SO2

d. PCl5 h. ClBr

3. Write the formulas for the following compounds.

a. Sulfur(II) chloride

b. Nitrogen(V) oxide

c. Carbon(IV) chloride

Names & Formulas for Acids

An acid is a distinct type of molecular compound. Most acids used in the laboratory can be classified as either binary acids or oxyacids. Binary acids are acids that consist of two elements, hydrogen and a nonmetal. Oxyacids are acids that contain hydrogen, oxygen, and a third element (usually a nonmetal). Oxyacids acids are derived from polyatomic ions.

To name a binary acid, follow this guideline.

Hydro – STEM – ic acid

Examples:

HCl hydrochloric acid

Hydrosulfuric acid H2S


To name an oxyacid, follow this guideline.

-ate –ic, -ite –ous

-ate ions make –ic acids and –ite ions make –ous acids

Examples:

H2SO4 sulfuric acid

H2SO3 sulfurous acid

HClO4 perchloric acid

Nitric acid HNO3

Nitrous acid HNO2

Carbonic acid H2CO3

Task 5j

1. Name the following acids.

a. H3PO4 c. HI

b. HBr d. HC2H3O2

2. Write the formulas for the following acids.

a. hypochlorous acid c. arsenic acid

b. hydroiodic acid d. hydroselenic acid

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Topic 6: Chemical Bonding & Molecular Geometry

Chemical Bonding (Chapter 6 in Modern Chemistry)

Atoms seldom exist as independent particles in nature. Most substances consist of combinations of atoms that are held together by chemical bonds. A chemical bond is a mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together. Why do atoms make bonds? It turns out that most atoms are less stable existing by themselves than when they are combined. By bonding with each other, atoms decrease in potential energy, thereby creating more stable arrangements of matter.

When atoms bond, their valence electrons are redistributed in way that make the atoms more stable. The way in which the electrons are redistributed determines the type of bonding. In Topic 4 you learned the main-group metals tend to lose electrons to form positive ions, or cations. Nonmetals tend to gain electrons to form negative ions, or anions. Chemical bonding that results from the electrical attraction between cations and anions is called ionic bonding. If a bond is purely ionic, an atom will completely give up electron(s) to another atom.

In contrast, atoms joined by covalent bonding share electrons. Covalent bonding results from the sharing of electron pairs between two atoms. In a purely covalent bond, the shared electrons are “owned” equally by the two bonded atoms.

an example of an ionic bond an example of a covalent bond

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Ionic or Covalent?

Bonding between atoms is rarely purely ionic or purely covalent. It usually falls somewhere between these two extremes, depending on how strongly the atoms of each element attract electrons. Recall that electronegativity is a measure of an atom’s ability to attract electrons. To determine whether a bond is ionic or covalent you have to calculate the difference in electronegativities of the two atoms involved.

An electronegativity difference greater than 1.67 is referred to as an ionic bond.

Electronegativity differences of 1.67 or less have an ionic bond character of 50% or less. These compounds are typically classified as covalent. Bonding between two atoms of the same element is completely covalent. This is called a nonpolar-covalent bond. This is a covalent bond in which the bonding electrons are shared equally by the bonded atoms, resulting in a balanced distribution of electrical charge. Bonds having only 0% to 5% ionic character, or an electronegativity difference equal to or less than 0.3, are considered nonpolar-covalent.

Bonds having an ionic character between 5% and 50%, or with corresponding electronegativity differences of 0.3 to 1.67, are classified as polar-covalent. A polar-covalent bond is a covalent bond in which the bonded atoms have an unequal attraction for the shared electrons.

Here are some example of determining bond character based on the electronegativities from the periodic table on the next page.

Bonding Pair Electronegativity

difference Bond type Li and F 3.98 - 0.98 = 3.00 ionic

Cu and S 2.58 – 1.90 = 0.68 polar-covalent

I and Br 2.96 -2.66 = 0.30 nonpolar-covalent

In summary, subtract the electronegativities, if:

Greater than 1.67 ionic Above 0.3 to 1.67 polar-covalent 0.3 or less nonpolar-covalent

3.3

Ionic

100%

1.67

Polar -covalent

50%

0.3 Nonpolar-covalent 5%

0

0%

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Electronegativities of the elements

Periodic table of electronegativity using the Pauling scale

Please note, if you do not have the electronegativities, ionic compounds are generally made up of elements that are far apart on the periodic table. For example: K & Cl make an ionic bond. Sr & Br make an ionic bond. S & O make a covalent bond. (Watch out for hydrogen; even though it is in group 1, it has a high electronegativity.)

Task 6a

1. Using the electronegativity chart, determine the type of bond between the atoms in the following compounds.

a. AgCl b. K2O c. Br2 d. HCl

→ Atomic radius decreases → Ionization energy increases → Electronegativity increases → Group

(vertical) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Period (horizontal)

1

H 2.20

He

2

Li 0.98

Be 1.57

B 2.04

C 2.55

N 3.04

O 3.44

F 3.98

Ne

3

Na 0.93

Mg 1.31

Al 1.61

Si 1.90

P 2.19

S 2.58

Cl 3.16

Ar

4

K 0.82

Ca 1.00

Sc 1.36

Ti 1.54

V 1.63

Cr 1.66

Mn 1.55

Fe 1.83

Co 1.88

Ni 1.91

Cu 1.90

Zn 1.65

Ga 1.81

Ge 2.01

As 2.18

Se 2.55

Br 2.96

Kr 3.00

5

Rb 0.82

Sr 0.95

Y 1.22

Zr 1.33

Nb 1.6

Mo 2.16

Tc 1.9

Ru 2.2

Rh 2.28

Pd 2.20

Ag 1.93

Cd 1.69

In 1.78

Sn 1.96

Sb 2.05

Te 2.1

I 2.66

Xe 2.60

6

Cs 0.79

Ba 0.89

*

Hf 1.3

Ta 1.5

W 2.36

Re 1.9

Os 2.2

Ir 2.20

Pt 2.28

Au 2.54

Hg 2.00

Tl 1.62

Pb 2.33

Bi 2.02

Po 2.0

At 2.2

Rn 2.2

7

Fr 0.7

Ra 0.9

**

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

Cn

Uut

Uuq

Uup

Uuh

Uus

Uuo

Lanthanoids

*

La 1.1

Ce 1.12

Pr 1.13

Nd 1.14

Pm 1.13

Sm 1.17

Eu 1.2

Gd 1.2

Tb 1.1

Dy 1.22

Ho 1.23

Er 1.24

Tm 1.25

Yb 1.1

Lu 1.27

Actinoids

**

Ac 1.1

Th 1.3

Pa 1.5

U 1.38

Np 1.36

Pu 1.28

Am 1.13

Cm 1.28

Bk 1.3

Cf 1.3

Es 1.3

Fm 1.3

Md 1.3

No 1.3

Lr 1.3

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2. Only using the periodic table, determine if the bonds below are covalent or ionic. Do not use the electronegativity chart.

a. XeCl6 b. CsF c. MgCl2 d. NO2

Ionic Bonding and Ionic Compounds

An ionic compound is composed of positive and negative ions that are combined so that the numbers of positive and negative charges are equal. The chemical formula of an ionic compound shows the ratio of ions present in a sample of any size. A formula unit is the simplest collection of atoms from which an ionic compound’s formula can be established. For example, one formula unit of sodium chloride, NaCl, is one sodium cation plus one chloride anion. (In the naming of a monatomic anion, the ending of the element’s name is replaced with –ide.) The ratio of ions in a formula unit depends on the charges of the ions combined. They must achieve electrical neutrality. In other words, the sum of the charges must equal zero.

For example: calcium fluoride

calcium has a 2+ charge, Ca2+ fluorine has a 1- charge, F-

In order for their charges to equal zero, there has to be two fluoride ions for every calcium ion.

CaF2

The Formation of Ionic Compounds Consider that a sodium atom and a chlorine atom are approaching each other. The two atoms are neutral and have one and seven valence electrons, respectively.

Sodium loses its electron to chlorine forming Na+ and Cl-.

Sodium atom + Chlorine atom Æ Sodium ion + Chloride ion

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Covalent Bonding

Most substances are composed of molecules. A molecule is a neutral group of atoms that are held together by covalent bonds. A molecule may consist of two of more atoms of the same element, as in oxygen, or of two or more different atoms, as in water or sugar.

Oxygen molecule, Water molecule, Sucrose molecule,

O2 H2O C12H22O11

A chemical compound whose simplest units are molecules is called a molecular compound. The composition of a compound is given by its chemical formula. A chemical formula indicates the relative numbers of atoms of each kind in a chemical compound by using atomic symbols and numerical subscripts. The chemical formula of a molecular compound is referred to as a molecular formula. A molecular formula shows the types and numbers of atoms combined in a single molecule of a molecular compound. The molecular formula for water, for example, is H2O, which reflects the fact that a single water molecule consists of one oxygen atom joined by separate covalent bonds to two hydrogen atoms. A molecule of oxygen, O2, is an example of a diatomic molecule. A diatomic molecule is a molecule containing only two atoms.

Formation of a Covalent Bond

Remember that nature favors chemical bonding because most atoms have lower potential energy when they are bonded to other atoms than they have when they are not bonded.

Using the example of two hydrogen atoms, if they are separated by enough distance, they will not influence each other. However, if the two hydrogen atoms approach each other, there will come to a position in which the nucleus of one hydrogen atoms attracts the electron from the other hydrogen atom. The attraction corresponds to a decrease in the total potential energy of the atoms. At the same time the electrons of the two hydrogen atoms are repelling each other. The protons in the nuclei of the two hydrogen atoms are also repelling each other. The repulsion results in an increase in potential energy. The relative strength of attraction and repulsion between the charged particles is dependent on the distance separating the two hydrogen atoms.

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Here the arrows indicate the attractive and repulsive forces between the electrons and nuclei of two hydrogen atoms. Attraction (red) between particles corresponds to a decrease in potential energy of the atoms, while repulsion (blue) corresponds to an increase.

The attractive force dominates and continues to pull the two hydrogen atoms closer together until they get to a distance at which the repulsion between like charges equals the attraction between opposite charges. At this position the two hydrogen atoms (now a hydrogen molecule, H2) have their minimum potential energy and are close enough to share electrons. They are now covalently bonded.

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Hydrogen Hydrogen atoms molecule

All individual hydrogen atoms contain a single, unpaired electron in a 1s atomic orbital. When two hydrogen atoms form a molecule, they share electrons in a covalent bond. The sharing of the electrons allows each atom to have the stable electron configuration of helium, 1s2. The tendency is for atoms to achieve a noble-gas configuration by sharing electrons. The Octet Rule

Unlike other atoms, the noble-gas atoms exist independently in nature. They possess a minimum of energy existing on their own because of the special stability of their electron configuration. This stability results from the fact that, with the exception of helium and its two electrons in a completely filled outer shell, the noble-gas atoms’ outer s and p orbitals are completely filled by a total of eight electrons.

Other main-group atoms can effectively fill their outermost s and p orbitals with electrons by sharing electrons through covalent bonding. Such bond formations follow the octet rule. Chemical compounds tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons init highest occupied energy level. Here is an example of the covalent bonding of hydrogen and fluorine to make hydrofluoric acid, HF. H H F F

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Exceptions to the Octet Rule

Most main-group elements tend to form covalent bonds according to the octet rule, However, there are exceptions. Hydrogen forms bonds in which there are only two electrons. Boron, B, and aluminum, Al, has just three valence electrons so it makes bonds surrounded by 6 electrons. Other atoms can be surrounded by more than eight electrons. Examples of elements that make bonds with expanded valences are phosphorus, P, in PF5 and sulfur, S, in SF6.

Task 6b

1. Use orbital notation to illustrate the bonding in the chlorine molecule, Cl2.

2. Describe the general location of the electrons in a covalent bond.

Metallic Bonding

Chemical bonding is different in metals than it is in ionic or molecular compounds. The highest energy levels of most metal atoms are occupied by very few electrons. The properties of metals are due to the highly mobile valence electrons of the atoms that make up a metal. The highest energy levels of most metal atoms are occupied by very few electrons. They have many vacant orbitals. The vacant orbitals in the atoms’ outer energy levels overlap. This overlapping of orbitals allows the outer electrons of the atoms to roam freely throughout the entire metal. The electrons are delocalized, which means that they do not belong to any one atom but move freely about the metal’s network of empty atomic orbitals. These mobile electrons form a sea of electrons around the metal atoms, which are packed together in a crystal lattice. The chemical bonding that results from the attraction between metal atoms and the surrounding sea of electrons is called metallic bonding.

Here are two diagrams to help you understand how the electrons surround the cation in a metallic bond.

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Van der Waals force

Van der Waals forces are a group of weak intermolecular forces that vary in strength but are generally weaker than bonds (ionic, covalent, and metallic). We will discuss hydrogen bonding, dipole-dipole interactions, and London dispersion forces (LDF).

Hydrogen Bonding

Some hydrogen-containing compounds, such as hydrogen fluoride, water, and ammonia, have unusually high boiling points. This is explained by the presence of a particularly strong type of dipole-dipole force. In compounds that contain hydrogen and a very electronegative element (N, O, F) an intermolecular attraction occurs between the molecules. Note that this is not a bond between atoms within the molecule, but an attraction among the molecules. Below, the dotted lines represent the hydrogen bonds between water molecules. This allows water to have relatively high melting and boiling points for hydrogen compounds that bond with group 16 nonmetals. This also explains why ice expands and floats.

Dipole-Dipole Attractions

A dipole is a molecule or a part of a molecule that contains both positively and negatively charged regions. For example, the molecule, HCl is made with hydrogen and the very electronegative atom chlorine. The electrons in this molecule tend to gather around the chlorine. This makes the hydrogen end more positively charged (G read as partially positive) and the chlorine end more negative (G-, read as partially negative). Each end is a dipole. H Cl

G+ G-


That means that when two hydrogen chloride molecules get close together and they are oriented correctly, the different dipoles will be attracted to each other.

London Dispersion Forces

There is some type of intermolecular force among all atoms and molecules, even noble gases. These are the weakest of the van der Waals forces and are called the London dispersion forces. London dispersion forces (LDF) are intermolecular attractions resulting from the constant motion of electrons and the creation of instantaneous dipoles. The more electrons there are the greater the London dispersion forces. So it goes to reason that the bigger the atomic mass or molecular mass the greater the London dispersion forces. That is why boiling points generally increase down a group on the periodic table.

Summary of Bonding and their relationship to properties

Ionic (Giant lattice)

+ve ion and -ve ion formed via transfer of electrons held together in giant lattice with

strong electrostatic interaction

Metal + Non-Metal ex: KCl, MgF2, Na2SO4, etc.

Strong Bonds

high m.p./b.p. poor conductors of

electricity when solid (ions not free to move), good

when liquid or in solution (dissolved)

will dissolve in polar solvents

Covalent (Individual molecules)

Small groups of atoms covalently bonded together by sharing electrons

Non-Metal + Non-Metal ex: CO2, PCl5, etc.

Strong bonds within molecules, but weak between molecules

low m.p/b.p., often liquids or gases at RT;

LDF- (induced dipoles) Dipole - perm. dipoles) H-Bonds - (H - N/O/F)

poor conductors

Metallic (Mixtures of Metals)

Close packed array of atoms (ions) with "sea" of free moving electrons

Metals ex: Na, Al, Au, Stainless Steel (Fe/C/Cr),

Bronze (Cu/Sn), Brass (Cu/Zn)

Strong (but flexible) bonds

high m.p/b.p. good conductors of electricity ("sea' of electrons) and heat (close packed), malleable (can be shaped), ductile (drawn into

thin wires), luster (shiny)


Task 6c

1. Compound B has lower melting and boiling points than compound A. At the same temperature, compound B vaporizes faster than compound A. If one of these compounds is ionic and the other is molecular, which would you expect to be molecular? Ionic? Explain your reasoning.

2. Analyzing Data. The melting points for the compounds Li2S, Rb2S, and K2S are 900oC, 530oC, and 840oC, respectively. List these three compounds in order of increasing lattice energy.

3. Explain why most metals are malleable and ductile but ionic crystals are not.

4. Explain why metals are good electrical conductors.

5. What is the difference between a formula unit and a molecule.

Lewis Structures

In Topic 3, you learned about electron dot diagrams for elements. These can be used to represent molecules also. When representing molecules, they are called Lewis structures. Lewis structures are formulas in which atomic symbols represent nuclei and inner-shell electrons, dot-pairs or dashes between two atomic symbols represent electron pairs in covalent bonds, and dots adjacent to only one atomic symbol represent unshared electrons. If only the shared pairs (bonds) are written using dashes, then this will be a structural formula. A structural formula indicates the kind, number, arrangement, and bonds but not the unshared pairs of the atoms in a molecule.

It is important to learn how to draw the Lewis structures of molecules and to predict the molecular geometry of the molecule. To predict the geometry, you have to consider shared (bonded) electrons and lone (nonbonded) electron pairs surrounding the central atom. To do this, we will use the VSEPR theory.

VSEPR stands for “valence shell electron pair repulsion. That means that the electron pairs around the atoms (usually the central atom) repel each other to affect the shape of the molecule. VSEPR theory states that the repulsion between the sets of valence-level electrons surrounding an atom causes these sets to be oriented as far apart as possible. Unshared electrons repel the most. Remember these molecules are 3-D even though we draw them in 2-D.

Drawing Lewis structures


Here are the steps for writing a Lewis structure for compounds for which you do not know the formula.

1. Calculate the total number of valence electrons taking into account any charges. (add electrons for negative charges and subtract electrons for positive charges)

2. Decide which element is the central atom. Usually this is obvious; if in doubt it will be the least electronegative atom (except for H: it only wants 2 electrons). Put that element in the center and add the other elements around the central atom using lines to represent bonding pairs of electrons.

3. Arrange the remaining electrons to complete the octet of the terminal atoms by placing pairs of dots around the atoms. Place any remaining electrons (dots) on the central atom, if necessary expanding the octet.

4. If the central atom lacks an octet, form multiple bonds (double or triple bonds) by converting non-bonding electrons on terminal atoms into bonding pairs. (Sometimes atoms will remain electron deficient).

For example:

CCl4 Valence electrons = 4 + 7(4) = 32

H2O Valence electrons = 1(2) + 6 = 8


Now practice with your teacher the following:

NH3

HF

PF5

NH4+

PCl6-

CO32-


Task 6d

1. Draw a Lewis structure for each of the following molecules. You will come back to this section for later tasks, so be sure to be neat and orderly.

a. SCl2

b. PI3

c. Cl2O

d. NH2Cl

e. SiCl3Br

f. ONCl

g. SO42-


h. ClO2-

i. BeCl2

You can also use VSEPR theory to predict the shapes of a molecule. Here is a link that might help you understand the shapes.

http://library.thinkquest.org/C006669/data/Chem/bonding/shapes.html

You will need to learn the chart on the next page.


Task 6e

1. Using the Lewis structures that you drew in 6d, determine the molecular geometry for each.

Hybridization

VSEPR theory helps determine the shapes of a molecule but it does not reveal the relationship between a molecule’s geometry and the orbitals occupied by its bonding electrons. To explain how the orbitals of an atom become rearranged when the atom forms covalent bonds, a different model is used. This model is called hybridization, which is the mixing of two or more atomic orbitals of similar energies on the same atom to produce new hybrid atomic orbitals of equal energies.

Methane, CH4, provides a good example of how hybridization is used to explain the geometry of molecular orbitals. The orbital notation for carbon’s valence electrons is 2s2 2p2. We know form experiments that a methane molecule has tetrahedral geometry. How does carbon form four equivalent, tetrahedrally arranged covalent bonds? The one s orbital and the three p orbital hybridize to form four new, identical orbitals called sp3 orbitals.

2p sp3

2s

Carbon’s orbitals Carbon’s orbitals after Before hybridization sp3 hybridization

Hybrid orbitals are orbitals of equal energy produced by the combination of two or more orbitals on the same atom. The number of hybrid orbitals produced equals the number of orbitals that have combined. A molecule with 3 bonding areas will be sp2, while a molecule with 4 bonding areas will be sp3. Trigonal bipyramidal molecules will be dsp3 and octahedral molecules will be d2sp3.


Polarity & Dipole forces The strongest intermolecular forces exist between polar molecules. Polar molecules act as tiny dipoles because of their uneven charge distribution. A dipole is created by equal but opposite charges that are separated by a short distance. The direction of the dipole is from the dipole’s positive pole to its negative pole. A dipole is represented by an arrow with a head pointing toward the negative pole and a crossed tail situated at the positive pole. The negative pole will be the atom that is the most electronegative. For example, hydrochloric acid, HCl:

H__Cl

This is a polar bond and a polar molecule because the charges are unevenly distributed. If the charge distribution is evenly distributed, the molecule will be nonpolar. For example CH4:

H

H __ C __ H

H

Notice that all the dipole point toward the C. Here the charge is evenly distributed. Therefore the molecule is nonpolar even though the individual bonds are polar.

Task 6f

1. Go back to Task 6d and label the hybridization and the polarity of each molecule.

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Topic 7: The Mole Concept Relating Mass to Numbers of Atoms (Chapter 3 in Modern Chemistry beginning on p.82)

In order to understand the quantitative parts of chemistry, there are three very important concepts—the mole, Avogadro’s number, and molar mass. These provide the basis for relating masses in grams to number of atoms. The Mole The mole is the SI unit for amount of substance. A mole (abbreviated mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. The mole is a counting unit, just like a dozen is. Avogadro’s number The number of particles in a mole has been experimentally determined in a number of ways. The best modern value is 6.022 1415 x 1023. This is the number of particles (atoms, ions, molecules, formula units) in a mole. It is called Avogadro’s number. Avogadro’s number—6.022 1415 x 1023—is the number of particles in exactly one mole of a pure substance. For most purposes, Avogadro’s number is rounded to 6.022 x 1023. Watch this video for an idea of how many particles Avogadro’s number really is. http://www.youtube.com/watch?v=1R7NiIum2TI Molar Mass An alternative definition of mole is the amount of a substance that contains Avogadro’s number of particles. The mass of one mole of a pure substance is called the molar mass of that substance. Molar mass is usually written in units of g/mol. The molar mass of an element is numerically equal to the atomic mass of the element in atomic mass units (which can be found in the periodic table).

Using Chemical Formulas (Chapter 7 in Modern Chemistry beginning on p.237)

As you have seen, a chemical formula indicates the elements as well as the relative number of atoms or ions of each element present in a compound. Chemical formulas also allow chemists to calculate a number of characteristic values for a given compound. In this section, you will learn how to use chemical formulas to calculate the molar mass, formula mass, and the percentage composition by mass of a compound.

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Molar Mass/Formula Mass Molecular and formula mass use the amu that we learned in Chapter 3. We use the atomic mass units from the periodic table. For example: Find the formula mass of potassium chlorate, KClO3. 1 K atom x 39.10 amu = 39.10 amu 1 Cl atom x 35.45 amu = 35.45 amu 3 O atoms x 16.00 amu = 48.00 amu Formula mass of KClO3 = 122.55 amu The term molecular mass is usually used for covalent compounds. The term formula mass is usually used for ionic compounds. Molar mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. The units for these calculations are g/mole. This is calculated the same way as molecular or formula mass. Example: Find the molar mass of Ba(NO3)2. 1 mole Ba x 137.33 g/mole = 137.33 g Ba 2 moles N x 14.01 g/mole = 28.02 g N 6 moles O x 16.00 g/mole = 96.00 g O Molar mass of 1 mole of Ba(NO3)2 = 261.35 g/ mole Task 7a 1. Find the molar mass of the following. a. NaCl b. K2O c. NaOH d. Ca(OH)2 e. (NH4)3PO4

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Percent Composition Sometimes you need to know the percentage by mass of a particular element in a chemical compound. For example, suppose the compound potassium chlorate, KClO3, were to be used as a source or oxygen. What is the percentage of oxygen in this compound? Use the following formula:

mass of element in sample of compound x 100 = % of element in compound Mass of sample of compound

The percentage by mass of each element in a compound is known as the percentage composition of the compound. Find the percentage composition of copper(I) sulfide, Cu2S. Find the molar mass of each element and of the compound. 2 Cu x 63.55 g/mol = 127.1 g/mole 1 S x 32.07 g/mol = 32.07 g/mole Molar mass of Cu2S = 159.2 g/mole Put these numbers into the formula for percent composition.

% Cu = 127.1 g/mol

x 100 = 79.85% 159.2 g/mole

% S =

32.07 g/mole x 100 = 20.15%

159.2 g/mol Task 7b 1. Find the percent composition of each element in the following compounds. (Note: You already calculated the molar mass in Task 7a). a. NaCl b. K2O c. NaOH d. Ca(OH)2 e. (NH4)3PO4

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Hydrates In some salts, water molecules from solution bind within their crystal structure. For example sodium carbonate forms a hydrate, in which 10 water molecules are present for every formula unit of sodium carbonate. The formula of the hydrate is Na2CO3

. 10 H2O. Note that the first part of the formula is the formula of the compound, then a dot. Following the dot, the number of water molecules that are attached to each formula unit is written. Hydrates are named by naming the compound, prefix hydrate. Again, using the example above; the hydrate is named sodium carbonate decahydrate. The molar mass of a hydrate is calculated by the following formula: MMhydrate = MMsalt + (# of water molecules x MMwater) Na2CO3

. 10 H2O 2(23) + 12 + 3(16) + 10[2(1) + 16] 106 + 10(18) 106 + 180 286 g/mol Task 7c 1. What is the name of MgSO4 . 5 H2O? 2. Write the formula of copper(II) chloride trihydrate. 3. Calculate the molar mass of CuSO4

. 5 H2O. 4. What is the percent of water in CuSO4 . 5 H2O? The Mole Why are M & Ms sold in a package rather than individually? List other items we package and then refer to as a package rather than the individual item: 100 pennies = 2 socks = 12 eggs =

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A chemist does the same thing with atoms! Why? A chemist’s package is a mole. 1 mole = 6.02 x 1023 particles. 6.02 x 1023 is called Avogadro’s number. It has been experimentally determined that when one measures out the mass of an element equal to its average atomic mass, the number of atoms contained in the sample is equal to 6.02 x 1023 atoms.

6.02 x 1023 atoms = 1 mole = Avogadro’s number = molar mass

Mole Conversions Particles to moles

o Convert representative particles to moles and moles to representative particles. (Representative particles are atoms, molecules, formula units, and ions.)

Mass to moles

o Convert mass of atoms, molecules, and compounds to moles and moles of atoms, molecules, and compounds to mass.

o Convert representative particles to mass and mass to representative particles. Volume of a gas to moles

o Convert moles to volume and volume to moles at STP.

Mass Moles Particles (grams) atoms, ions, molecules,

formula units

MM from periodic table Use 6.02 x 1023

(g/mol) Volume of Gas (at STP) (particles/mole) (liters)

22.4 L/mole

The mole is the central unit in all mole conversions. The other units are used as needed for conversion factors.

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The key to these problems is using dimensional analysis. Remember the key to dimensional analysis is to cancel out the units you do not need and keep the units you do need. Summary If grams are in the problem: molar mass from the periodic table g/mole If particles are in the problem: Avogadro’s number 6.02 x 1023 particles/mole If liters of gas are in the problem: molar volume 22.4 L/mole Examples: Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. 1) If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle? 2) How many molecules of ibuprofen are in the bottle? 3) What is the total mass in grams of carbon in 33 g of ibuprofen? Answers 1) 33 g C13H18O2 1 mole C13H18O2 = 0.16 mole C13H18O2 206.31 g C13H18O2 2) 0.16 mole C13H18O2 6.02 x 1023 molecules = 9.6 x1022 molecules C13H18O2 mole 3) 0.16 mole C13H18O2 13 moles of C 12.01 g C = 25 g C 1 mole C13H18O2 mole Task 7d

1. How many atoms in 400.0g of sulfur?

2. What is the mass of 1.2 x 1024 atoms of magnesium?

3. What is the mass of 2.5 moles of oxygen atoms?

4. Given 18 grams of water, how many molecules do you have?

5. What is the mass, in grams, of 1.2 x 1024 formula units of sodium chloride?

6. What is the volume of carbon dioxide gas in 589.4 g of carbon dioxide?

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Empirical & Molecular Formulas Empirical Formula – the formula for a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula. Molecular Formula- the formula for a compound in which the subscripts give the actual number of each element in the formulas it truly exists. An empirical formula cannot be reduced any further. A molecular formula may or may not be reduced.

Molecular Formula Empirical Formula H2O

CH3COOH

CH2O

C6H12O6

Notice two things: 1. The molecular formula and the empirical formula can be identical.

2. You scale up from the empirical formula to the molecular formula by a whole number factor.

Calculating Empirical Formulas A Simple Rhyme for a Simple Formula by Joel S. Thompson

Percent to mass Mass to mole

Divide by small Multiply ‘til whole

In other words:

1. Since percent can be on any mass, assume a 100 g sample and change directly to mass.

43 % of 100 g is 43 g. 2. Use molar mass to perform a mole conversion to change gram to moles. 3. Divide by the smallest number of moles to try to get a whole number ratio.

4. If you do not get a whole number, multiply by a whole number until you get a whole number. Remember formulas cannot be in fractions.

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Example #1 Quantitative analysis shows that a compound contains 32.38 % sodium, 22.65 % sulfur, and 44.99 % oxygen. Find the empirical formula of this compound. Change % Use MM to change Divide by the Hopefully, get a to grams g to moles smallest moles whole number 32.38 g Na mole Na = 1.408 mole Na = 2 22.99 g 0.7063 mole Write the formula 22.65 g S mole S = 0.7063 mole S = 1 Na2SO4 32.07 g 0.7063 mole 44.99 g O mole O = 2.812 mole O = 4 This is the 16.00 g 0.7063 mole empirical formula Example #2 Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this compound? Do not have to Change g Divide by the Didn’t get a whole change to grams, to moles smallest number for both, already in grams so multiply by a number to get a whole number 4.433 g P mole P = 0.1431 mole P = 1 (2) = 2 Write the 30.97 g P 0.1431 mole formula P2O5 5.717 g O mole O = 0.3573 mole O = 2.5 (2) = 5 16.00 g O 0.1431 mole The grams of O came from subtracting 10.150 g – 4.433 g

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Task 7e 1. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? 2. A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of the compound is known to be approximately 140 g/mol. What is the empirical formula? 3. A compound is found to contain 63.52 % iron and 36.48 % sulfur. Find its empirical formula. 4. Find the empirical formula of a compound found to contain 26.56 % potassium, 35.41 % chromium, and the remainder oxygen.

Calculating Molecular Formulas To determine the molecular formula you must know the empirical formula and the molecular mass. Remember the molecular formula is a multiple of the empirical formula. To find the multiple: molar mass of the molecular formula = x molar mass of the empirical formula To write the molecular formula: multiply the subscripts in the empirical formula by x Example #3 In example problem #2, the empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? Molecular formula’s mass = 283.89 g/mol Empirical formula’s mass = 141.94 g/mol 283.89 g/mol = 2 P2O5 x 2 = P4O10 141.94 g/mol

The empirical formula was given in the problem.


Example #4 A compound with a formula mass of 42.08 amu is found to be 85.64 % carbon and 14.36% hydrogen by mass. Find the molecular formula. Find the empirical formula first. 85.64 g C mole C = 7.131 mol C = 1 Empirical formula 12.01 g C 7.131 mol CH2 14.36 g H mole H = 14.25 mol H = 2 1.008 g H 7.131 mol 42.08 amu = 3 CH2 (3) = C3H6 14.03 amu Molecular formula

Empirical formula mass

Task 7f 1. A 1.50 g sample of hydrocarbon undergoes complete combustion to produce CO2 and H2O. The empirical formula of this compound is CH3. Its molecular weight has been determined to be about 78. What is the molecular formula? 2. Caffeine has the following percent composition: carbon 49.48%, hydrogen 5.19%, oxygen 16.48% and nitrogen 28.85%. Its molecular weight is 194.19 g/mol. What is its molecular formula?

Concentration of Solutions (Molarity) (Chapter 12 in Modern Chemistry beginning on p.418)

The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. In this section, we introduce two different ways of expressing the concentrations of solutions: molarity and molality. Sometimes, solutions are referred to as “dilute” or “concentrated,” but these are not very definite terms. “Dilute” just means that there is a relatively small amount of solute in a solvent. “Concentrated,” on the other hand, means that there is a relatively large amount of solute in a solvent. These terms are unrelated to the degree to which a solution is saturated. A saturated solution of a substance that is not very soluble might be very dilute.


Molarity Molarity is the number of moles of solute in one liter of solution. Molarity is symbolized with a “M” or brackets, [ ]. The formula for finding molarity is:

molarity (M) = moles of solute (mol) volume of solution (L)

Note that the solute must be in moles and the volume of solution must be in liters.

It is also important to realize that making a 1 molar (1 M) solution is not made by adding 1 mol of solute to 1 L of solvent. You need to dissolve the solute in a small amount of water then add enough water to make the total volume of solution 1L. Here are some examples of molarity problems. Ex. 1 You have 3.50 L of solution that contains 90.0 g of sodium chloride. What is the molarity of this solution?


First you have to change the 90.0 g of sodium chloride to moles of sodium chloride.

90.0 g NaCl moles of NaCl = 1.54 mol NaCl

58.5 g of NaCl

Then plug this information into the molarity formula.

M = 1.54 mol NaCl = 0.440 M NaCl

3.50 L solution

OR You can do the same problem using dimensional analysis.

90.0 g NaCl moles of NaCl = 0.440 M NaCl

58.5 g of NaCl 3.50 L solution


Ex. 2 You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Remember that the capital “M” means moles of solute per liters of solution so 0.5 M means 0.5 moles of HCl in 1 liter of solution. Writing the molarity this way allows you to see that you can use dimensional analysis to cancel liters and solve for moles.

0.5 moles HCl 0.8 L solution = 0.4 mol HCl

L solution

You could also find the grams of HCl by using molar mass if needed. Ex. 3 To produce 40.0 g of silver chromate, you will need at least 23.4 g of potassium chromate in solution as a reactant. All you have on hand is 5 L of 6.0 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction? First, it is important to realize that the 40.0 g of silver chromate is not important information. We need to change 23.4 grams to moles.

23.4 g K2CrO4 1 mol K2CrO4 = 0.120 mol K2CrO4

194.2 g K2CrO4

Now plug the information into the molarity formula.


6.0 moles K2CrO4 = 0.120 mol K2CrO4

L solution x

x = 0.020 L K2CrO4 soln

OR You can do the same problem using dimensional analysis.

23.4 g K2CrO4 moles of K2CrO4 L of soln = 0.020 L K2CrO4 soln

194.2 g of K2CrO4 6.0 mol K2CrO4


Task 7g 1. What is the molarity of a solution composed of 5.85 g potassium iodide, KI, dissolved in enough water to make 0.125 L of solution? 2. How many moles of H2SO4 are present in 0.500 L of a 0.150 M H2SO4 solution? 3. What volume of 3.00 M NaCl is needed for a reaction that requires 146.3 g of NaCl? Molality Molality is the number of moles of solute per kilogram of solvent. Molality is symbolized with a “m”. The formula for finding molality is:

molality (m) = moles of solute (mol) Kilograms of solvent (kg)

Note that the solute must be in moles and the solvent must be in kilograms.

It is also important to realize that making a 1 molal (1 m) solution is made by adding 1 mol of solute to 1 kg of solvent. Here are some examples of molality problems. Ex. 1 A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C12H22O11) in 125 g of water. Find the molal concentration of this solution.

molality (m) = moles of solute (mol) Kilograms of solvent (kg)

First you have to change the 17.1 g of sucrose to moles of sucrose.

17.1 g C12H22O11 moles of C12H22O11 = 0.0500 mol C12H22O11

342.34 g of C12H22O11

Then plug this information into the molality formula, remembering that you have to use kg of solvent. 125 g of water is 0.125 kg of water.

m = 0.0500 mol C12H22O11 = 0.400 m C12H22O11 0.125 kg solvent

OR


You can do the same problem using dimensional analysis.

17.1 g C12H22O11 moles of C12H22O11 = 0.400 m C12H22O11

342.34 g of C12H22O11 0.125 kg solvent

Ex. 2 A solution of iodine, I2, in carbon tetrachloride, CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.480 m solution of iodine in CCl4 if 100.0 g of CCl4 is used? Remember that the “m” means moles of solute per kilograms of solvent so 0.480 m means 0.480 moles of I2 in 1 kg of CCl4. Writing the molality this way allows you to see that you can use dimensional analysis to cancel kilograms and solve for moles.

0.480 moles I2 0.100 kg solvent = 0.480 mol I2 kg solvent

It is then possible to change moles of iodine to grams of iodine by using the molar mass of iodine.

0.480 mole I2 253.8 g I2 = 12.2 g I2

mol I2

OR

You can do the same problem using dimensional analysis.

0.480 moles I2 0.100 kg solvent 253.8 g I2 = 12.2 g I2 kg solvent

mol I2

Task 7h 1. What is the molality of acetone in a solution composed of 255 g of acetone, (CH3)2CO, dissolved in 200. g of water? 2. What quantity, in grams, of methanol, CH3OH, is required to prepare a 0.244 m solution in 400. g of water?


Dilution Sometimes you will need to dilute a more concentrated solution. The formula for dilution is:

M1V1 = M2V2 For example: What volume of 6.0 M KCl will be needed to make 300.0 mL of 2.4 M KCl?

(6.0 M)V1 = (2.4 M) (300.0 mL) V1 = 120 mL

Task 7i 1. How would you make 300.0 ml of 2.4 M KCl from 6.0 M KCl?

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Topic 8: Chemical Reactions Chemical Equations & Reactions (Chapter 8 in Modern Chemistry)

Describing Chemical Reactions

A chemical reaction is the process by which one or more substances are changed into one or more different substances. In any chemical reaction, the original substances are known as the reactants and the resulting substances are known as the products. According to the law of conservation of mass, the total mass of reactants must equal the total mass of products for any given chemical reactions. Chemical reactions are described by chemical equations. A chemical equation represents, with symbols and formulas, the identities and relative molecular or molar amounts of the reactants and products in a chemical reaction. For example, the following chemical equation shows that the reactant ammonium dichromate yields the products nitrogen, chromium(III) oxide, and water.

(NH4)2Cr2O7(s) Æ N2(g) + Cr2O3(s) + 4 H2O(g)

Reactant Products Indications of a Chemical Reaction Since a chemical reaction produces new substances, the only way to determine if a chemical reaction has actually occurred is by chemical analysis of the product. However, certain easily observed changes usually indicate that a chemical reaction has occurred.

1. Evolution of energy as heat and light. A change in matter that releases energy as both heat and light is strong evidence that a chemical reaction has taken place. Some reactions involve only heat or only light. But heat or light by itself is not necessarily a sign of chemical change.

2. Production of a gas. The evolution of gas bubbles when two substances are mixed is often evidence of a chemical reaction.

3. Formation of a precipitate. Many chemical reactions take place between substances that are dissolved in liquids. If a solid appears after two solutions are mixed, a reaction has likely occurred. A solid that is produced as a result of a chemical reaction in solution and that separates from the solution is known as a precipitate.

4. Color change. A change in color is often an indication of a chemical reaction.

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Characteristics of Chemical Equations A properly written chemical equation can summarize any chemical change. Here is an example equation and a list of requirements that will aid you in writing and reading chemical equations correctly.

(NH4)2Cr2O7(s) Æ N2(g) + Cr2O3(s) + 4 H2O(g)

Subscripts physical state symbols Coefficient

“yields” or “produces”

1. The equation must represent know facts. All reactants and products must be identified. 2. The equation must contain the correct formulas for the reactants and products. The

symbols and formulas must be correct including elements, diatomic molecules (see Table 1), and ions.

3. The law of conservation of mass must be satisfied. Atoms can neither created nor destroyed in ordinary chemical reactions. So there must be the same number of each atom of each element on both side of the equation. To balance numbers of atoms, add coefficients where necessary. A coefficient is a small whole number that appears in front of a formula in a chemical equation. Placing a coefficient in front of a formula specifies the relative number of moles of the substance; if no coefficient is written the coefficient is understood to be 1.

Table 1 Elements that normally exist as Diatomic Molecules

Element Symbol Molecular Formula

Physical state at room temperature

Hydrogen H H2 gas Nitrogen N N2 gas Oxygen O O2 gas Fluorine F F2 gas Chlorine Cl Cl2 gas Bromine Br Br2 liquid Iodine I I2 solid

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Word and Formula Equations The first step in writing a chemical equation is to identify the facts to be represented. Information may be given in the form of a word equation or you may need to write a word equation. A word equation is an equation in which the reactants and products in a chemical reaction are represented by words. A word equation has only qualitative (descriptive) meaning. For example the word equation for the reaction of methane and oxygen is written as follows.

Methane + oxygen Æ carbon dioxide + water This equation would read: methane reacts with oxygen to produce carbon dioxide and water. Note that the arrow,Æ, could also be read as “yields” or “forms”. The next step in writing a correct chemical equation is to replace the names of the reactant and products with appropriate symbols and formulas. You should already know the formulas for most of the compounds that we will use. This formula equation represents the reactants and products of a chemical reaction by their symbols or formulas.

CH4 + O2 Æ CO2 + H2O (not balanced)

Additionally, it is appropriate to include the physical state symbols of the substance if known.

CH4(g) + O2(g) Æ CO2(g) + H2O(g) (not balanced)

The (g) after each formula indicates that the corresponding substance is in the gaseous state. You could also use (l) for liquid, (s) for solid, and (aq) for aqueous or water solution. Let’s practice writing formula equations from word equations. Task 8a Write chemical equation for each of the following sentences. Assume that these reactions take place at room temperature

1. Aluminum reacts with oxygen gas to produce solid aluminum oxide.

2. Phosphoric acid is produced through the reaction between tetraphosphorus decoxide and water.

3. Solid iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide.

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Formula equations meet two of the three requirements for a correct chemical equation. It represents the facts and shows the correct symbols and formulas for the reactants and products. It does not satisfy the law of conservation of mass. To complete this step, we must balance the equation. To balance an equation there must be the same number of atoms of each element on each side of the arrow. This is usually done by trial and error.

CH4(g) + O2(g) Æ CO2(g) + H2O(g) (not balanced) To balance equation, we use coefficients. Remember coefficients go with the entire compound. 2 CH4 would mean 2 carbon atoms and 8 hydrogen atoms. 2 Na2SO4 would mean 4 sodium atoms, 2 sulfur atoms, and 8 oxygen atoms. 3 (NH4)2CO3 would mean 6 nitrogen atoms, 24 hydrogen atoms, 3 carbon atoms, and 9 oxygen atoms. Task 8b

1. How many atoms of each type are represented in each of the following?

1. 3 N2

2. 2 H2O

3. 4 HNO3

4. 2 Ca(OH)2

5. 3 Ba(ClO3)2

6. 4 Mg3(PO4)2

7. 6 Al2(SeO4)3

8. 4 C3H8

Balancing Equations General balancing: Notice that you do not need the physical state symbols to balance the equation. They have no effect on the number of atoms so I will ignore them while balancing.

CH4(g) + O2(g) Æ CO2(g) + H2O(g) (not balanced) In this equation, the is 1 C on each side, so the carbons are balanced but there are 4 H’s on the reactant side and only 2 H’s on the product side. To balance this we will place a coefficient of 2 in front of the H2O. Never change subscripts when balancing an equation.

CH4(g) + O2(g) Æ CO2(g) + 2 H2O(g) (not balanced)

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Now there are 2 O’s on the reactant side of the equation but 4 O’s on the product side. (Notice that I added the 2 O’s from CO2 to the 2 O’s in H2O.) To balance this we will place a coefficient of 2 in front of the O2.

CH4(g) + 2 O2(g) Æ CO2(g) + 2 H2O(g) (balanced)

You could read this as: 1 mole of methane gas reacts with 2 moles of oxygen gas to produce 1 mole of carbon dioxide gas and 2 moles of water vapor.

It is always a good idea to recheck the equation after you are finished. Sometimes while changing the coefficient for one substance you may change one of the elements that you have already balanced. Also when there is no coefficient, it is understood to be 1. This is a simple equation. Balancing equations takes practice. The more you practice, the easier balancing will become. Equations containing polyatomic ions: If you notice that polyatomic ions are in the equation, and they do not split apart. It is sometimes easier to consider them as one entity.

H2SO4(aq) + NaNO3(aq) Æ HNO3(aq) + Na2SO4(aq) (not balanced) I would balance this by balancing the hydrogen atoms first. This requires me to put a 2 in front of HNO3.

H2SO4(aq) + NaNO3(aq) Æ 2 HNO3(aq) + Na2SO4(aq) (not balanced)

Now I have 2 H’s on each side. I also have 1 sulfate ion (SO4) on each side. Since sulfate stayed together, I do not have to treat the sulfur and oxygen atoms separately. This makes balancing easier since there are oxygen atoms in every substance. I have 1 Na atom on the reactant side but 2 Na atoms on the product side. I will place a 2 in front of NaNO3. This also gives me 2 nitrate ions (NO3) on each side.

H2SO4(aq) + 2 NaNO3(aq) Æ 2 HNO3(aq) + Na2SO4(aq) (balanced)

Equations containing water: Water can be written as H2O or HOH. Pick whichever is easier for you to balance.

HCl(aq) + NaOH(aq) Æ NaCl(aq) + HOH(l) (balanced) Here water is written as HOH because it is easier to see what happened and to balance the equation. This equation is already balanced as written. If you are given one of these in homework or on a test, I ask that you write “balanced” or put a “B” beside it. This tells me you knew what you were doing and not just skipping the problem.

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Task 8c Balance the following equations.

1. ____NaNO3 + ____PbO Æ ____Pb(NO3)2 + ____Na2O

2. ____AgI + ____Fe2(CO3)3 Æ ____FeI3 + ____Ag2CO3

3. ____C2H4O2 + ____O2 Æ ____CO2 + ____H2O

4. ____ZnSO4 + ____Li2CO3 Æ ____ZnCO3 + ____Li2SO4

5. ____V2O5 + ____CaS Æ ____CaO + ____V2S5

6. ____Mn(NO2)2 + ____BeCl2 Æ ____Be(NO2)2 + ____MnCl2

7. ____AgBr + ____GaPO4 Æ ____Ag3PO4 + ____GaBr3

8. ____H2SO4 + ____B(OH)3 Æ ____B2(SO4)3 + ____H2O

9. ____S8 + ____O2 Æ ____SO2

10. ____Fe + ____AgNO3 Æ ____Fe(NO3)2 + ____Ag

Task 8d

1. Write balanced equations for the following reactions. Include physical state symbols where indicated.

a. Solid calcium reacts with solid sulfur to produce solid calcium sulfide.

b. Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas.

c. Solid aluminum metal reacts with aqueous zinc chloride to produce solid zinc metal and aqueous aluminum chloride.

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2. Translate the following chemical equations into sentences:

a. CS2(l) + 3 O2(g) Æ CO2(g) + 2 SO2(g)

b. NaCl(aq) + AgNO3(aq) Æ NaNO3(aq) + AgCl(s)

Additional Symbols Used in Chemical Equations

Symbols Used in Chemical Equations

"Yields"; indicates result of reaction

Used in place of a single arrow to indicate a reversible reaction

(s) Solid state; also used to indicate a precipitate Used only to indicate a precipitate

(l) Liquid state

(aq) Aqueous solution; dissolved in water

(g)

Gaseous state

Used only to indicate a gaseous product

Reactants are heated

heat Reactants are heated

2 atm Pressure at which reaction is carried out, in this case 2 atm

pressure Indicates that reaction is carried out abouve normal atmospheric pressure

OoC Temperature at which reaction is carried out, in this case 0oC

MnO2 Formula of catalyst, used to alter the rate of the reaction

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Significance of a Chemical Equation Now that the equations are balanced, it is useful in doing quantitative chemistry.

1. The coefficients of a chemical reaction indicate relative, not absolute amounts of reactants and products. This is a ratio of the smallest numbers of atoms, molecules, or ions that will satisfy the law of conservation of mass.

H2(g) + Cl2(g) Æ 2 HCl(g)

This shows that 1 molecule of hydrogen gas reacts with 1 molecule of chlorine gas to produce 2 molecules of hydrogen chloride gas.

It could also represent 20 molecules of hydrogen gas reacts with 20 molecules of chlorine

gas to produce 40 molecules of hydrogen chloride gas.

Or in terms of amounts in moles: 1 mole of hydrogen gas reacts with 1 mole of chlorine gas to produce 2 mole of hydrogen chloride gas.

2. The relative masses of the reactants and products of a chemical reaction can be determined from the reaction’s coefficients. Remember in Topic 7 that you can convert moles to mass in grams by multiplying by the appropriate molar mass.

1 mol H2 x 2.02 g H2 = 2.02 g H2 mol H2

1 mol Cl2 x 70.90 g Cl2 = 70.90 g Cl2 mol Cl2

2 mol HCl x 36.46 g HCl = 72.92 g HCl

mol HCl

The chemical equation shows that the mass of the reactants: 2.02 g H2 + 70.90 g Cl2 equals the mass of the product: 72.92 g HCl

3. The reverse reaction for a chemical equation has the same relative amounts of substances as the forward reaction. This means that 2 moles of hydrogen chloride gas would decompose to make 1 molecule of hydrogen gas and 1 mole of chlorine gas.

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Types of Chemical Reactions There are thousands of known chemical reactions that occur. Often it is necessary to predict the products formed in one of these reactions. Therefore it is useful to classify reactions according to various similarities and regularities. There are several ways to classify chemical reactions and none are perfect but we will look at five basic types of reactions: synthesis, decomposition, single displacement, double displacement, and combustion. Later we will look at other types. Remember: 1. Write the symbols. 2. Check for diatomic molecules. 3. Check oxidation numbers. 4. Write physical state symbols if available. 5. Balance. The order of steps 1-4 can be mixed, but step 5 must be completed last. Synthesis In a synthesis reaction, two or more substances combine to form a new compound. General equation:

A + X Æ AX Example: Magnesium reacts with oxygen

Mg(s) + O2(g) Æ 2 MgO(s)

Oxides of metals are usually solids while nonmetallic oxides are usually

gases.

Oxygen is a diatomic molecule hence the

subscript 2 Note that to get the formula for magnesium

oxide oxidation numbers were used as in Topic 5. Mg has a 2+ charge and O has a 2- charge. Since they equal zero there are no

subscripts.


Special synthesis reactions

1. Metallic oxides react with water to produce metallic hydroxides.

CaO(s) + H2O(l) Æ Ca(OH)2(s)

2. Nonmetallic oxides react with water to produce oxyacids.

SO2(g) + H2O(l) Æ H2SO3(aq)

3. Metallic oxides can react with carbon dioxide to produce metallic carbonates.

Na2O(s) + CO2(g) Æ Na2CO3(s)

4. Metallic oxides can react with nonmetallic oxides to produce salts.

CaO(s) + SO2(g) Æ CaSO3(s)

Note that the oxidation number of the metal and the nonmetal will remain the same. This will allow you to determine the number of oxygen atoms in the product.

5. Metallic chlorides react with oxygen to produce metallic chlorates.

2 KCl(s) + 3 O2(g) Æ 2 KClO3(s)

Task 8e Write the balanced equations for the following synthesis reactions.

1. Lithium metal reacts with oxygen gas

2. Iron metal reacts with oxygen gas to produce an iron(III) compound

3. Calcium reacts with iodine

4. Barium oxide reacts with carbon dioxide


Decomposition In a decomposition reaction, a single compound undergoes a reaction that produces two or more simpler substances. . General equation:

AX Æ A + X Examples: Binary Compounds 1. Water decomposes by an electric current

The decomposition of a substance by an electric current is called electrolysis.

2 H2O(l) Æ 2 H2(g) + O2(g)

2. Mercury(II) oxide decomposes when heated.

Oxides of less active metals decompose into their elements when heated.

2 HgO(s) Æ 2 Hg(l) + O2(g)

Again, hydrogen and oxygen are diatomic so they have a subscript of 2 when they are by themselves.

You shouldn’t have to check the oxidation number of water. You should know that formula.

The lightning bolt represents electricity

Recall that the means heated.


Special decomposition reactions These reactions are the reverse of the synthesis reactions.

1. Metallic hydroxides decompose to produce metallic oxides and water.

NaOH(s) Æ Na2O(s) + H2O(s)

2. Oxyacids decompose into nonmetallic oxides and water.

H2SO4(aq) Æ SO3(g) + H2O(l)

3. Metallic carbonates decompose to metallic oxides and carbon dioxide.

CaCO3(s) Æ CaO(s) + CO2(g)

4. Metallic chlorates decompose to metallic chlorides and oxygen.

2 KClO3(s) Æ 2 KCl(s) + 3 O2(g)

Task 8f Write the balanced equations for the following decomposition reactions.

1. The decomposition of solid sodium chloride when heated

2. Solid calcium hydroxide decomposes

3. Solid silver chloride decomposes

4. Heat is applied to solid magnesium chlorate


Single Displacement In a single displacement reaction one element replaces a similar element in a compound. . General equation:

A + BX Æ AX + B

or

Y + BX Æ BY + X Examples: Aluminum metal reacts with a solution of lead(II)nitrate.

2 Al(s) + 3 Pb(NO3)2(aq) Æ 3 Pb(s) + 2 Al(NO3)3(aq) Chlorine gas reacts with potassium bromide.

Cl2(g) + 2 KBr(aq) Æ 2 KCl(aq) + Br2(l)

An activity series is a list of elements organized according to the ease with which the elements undergo certain chemical reaction. When working with single displacement reactions, it is important for you to check the activity. An activity series is located on the next page.

Here element A makes a positive ion so it reacts with the negative ion, X. It replaces B in the compound and B is the leftover element.

In this example, element Y makes a negative ion so it reacts with the positive ion, B. It replaces X in the compound BX and X is now the leftover element.


Activity Series of Metals Activity Series of Halogen Nonmetals

Li F2 Rb Cl2 K Br2 Ba I2 Sr Ca Na

Mg Al Mn Zn Cr Fe Cd

Co Ni Sn Pb

H2 Sb Bi Cu Hg

Ag Pt Au

Elements that are higher on the activity series will replace those that are below it. For example:

Will aluminum react with zinc chloride? Aluminum is higher on the activity series than zinc so yes it will react.

2 Al(s) + 3 ZnCl2(aq) Æ 3 Zn(s) + 2 AlCl3(aq)

Will cobalt react with sodium chloride? Cobalt is lower on the chart than sodium, therefore there is no reaction.

Co(s) + 2 NaCl(aq) Æ NR

React with cold H2O and acids, replacing hydrogen.

React with oxygen, forming oxides.

React with steam (but not with cold water) and acids, replacing hydrogen. React

with oxygen, forming oxides

Do not react with water. React with acids, replacing

hydrogen. React with oxygen, forming oxides

React with oxygen, forming oxides.

Fairly unreactive, forming oxides only indirectly.

“NR” stands for no reaction


You can also use the activity series for halogens by using the right side of the activity series. For example:

Will chlorine reacts with sodium fluoride? No, chlorine is lower on the activity series than fluorine.

Cl2(g) + NaF(aq) Æ NR

Task 8g Write the balanced equations for the following single displacement reactions (Assume all of these reactions react).

1. Zinc metal reacts with a lead(II) nitrate solution

2. Aluminum combines with a mercury(II) acetate solution

3. Aluminum reacts with a nickel(II) sulfate solution

4. Sodium metal reacts with water at room temperature Based on the activity series, predict whether each of the following reactions will occur. Write an appropriate equation for each.

1. Nickel reacts with water

2. Bromine reacts with potassium iodide

3. Gold reacts with hydrochloric acid

4. Cadmium reacts with hydrochloric acid

5. Magnesium reacts with cobalt(II) nitrate


Double Displacement In double displacement reactions, the ions of two compounds exchange places in an aqueous solution to form two new compounds. General equation:

AX + BY Æ AY + BX

Notice that the ions just switched partners. A combined with Y. B combined with X. Examples: An aluminum chloride reacts with a solution of lead(II)nitrate.

2 AlCl3(aq) + 3 Pb(NO3)2(aq) Æ 3 PbCl2(s) + 2 Al(NO3)3(aq)

Later in this topic you will learn how to determine the physical state symbols of reactions similar to this one. Task 8h Write the balanced equations for the following double displacement reactions.

1. A silver nitrate solution reacts with a solution of sodium chloride

2. Solutions of magnesium nitrate and potassium hydroxide combine

3. Hydrochloric acid reacts with sodium hydroxide in water

4. Copper(I) chloride reacts with barium fluoride

A is a positive ion. X is a negative ion.

B is a positive ion. Y is a negative ion.

Positive ion A combines with negative ion Y. Positive ion B combines with negative ion X.


Combustion In a combustion reaction, a substance combines with oxygen, releasing a large amount of energy in the form of light and heat. If the substance is a hydrocarbon, carbon dioxide and water will be produced in addition to energy. General equation:

Element + O2(g) Æ Oxide of element

Hydrocarbon + O2(g) Æ CO2(g) + H2O(l or g)

Examples: Carbon monoxide reacts with oxygen

2 CO(g) + O2(g) Æ 2 CO2(g)

The combustion of propane

C3H8(g) + 5 O2(g) Æ 3 CO2(g) + 4 H2O(g)

Task 8i Write the balanced equations for the following combustion reactions.

1. The combustion of ethane

2. The combustion of pentane

3. The reaction of strontium metal and oxygen

Remember, as long as the combustion is of a hydrocarbon, it will always be O2 Æ CO2 + H2O.

Only the balancing will change.

This part of the equation is always the same except for balancing.

The physical state symbol of water depends on the temperature.


Task 8j

1. Classify each of the following reactions as a synthesis, decomposition, single displacement, double displacement, or combustion reaction.

a. N2(g) + 3 H2(g) Æ 2 NH3(g)

b. 2 Li(s) + 2 H2O(l) Æ 2 LiOH(aq) + H2(g)

c. 2 NaNO3(s) Æ 2 NaNO2(s) + O2(g)

d. 2 C6H14(l) + 19 O2(g) Æ 12 CO2(g) + 14 H2O(l)

e. AgNO3(aq) + NaCl(aq) Æ AgCl(s) + NaNO3(aq)

2. Predict the products of the following reactions:

a. Silver reacts with copper(II) sulfate

b. Sodium iodide reacts with calcium chloride

c. Oxygen reacts with hydrogen

d. Nitric acid reacts with manganese(II) hydroxide

e. Silver nitrite reacts with barium sulfate

f. Hydrocyanic acid reacts with copper(II) sulfate

g. Water reacts with silver iodide


h. Nitric acid reacts with iron(III) hydroxide

i. Lithium bromide reacts with cobalt(II) sulfite

j. Lithium nitrate reacts with silver

k. Nitrogen gas reacts with oxygen gas

l. Carbonic acid decomposes

m. Aluminum chloride reacts with cesium

n. Aluminum nitrate reacts with gallium

o. Sulfuric acid reacts with ammonium hydroxide

p. Acetic acid reacts with oxygen gas

q. Butane burns in the presence of oxygen

r. Potassium chloride reacts with magnesium hydroxide

s. Zinc reacts with gold(II) nitrate


Ions in Aqueous Solutions (Chapter 13 in Modern Chemistry)

Dissociation When a compound that is made of ions dissolves in water, the ions separate from one another. This separation of ions that occurs when an ionic compound dissolves is called dissociation. For example, dissociation of sodium chloride and calcium chloride in water can be represented by the following equations.

NaCl(s) H2O > Na+(aq) + Cl-(aq) 1mole 1mole 1mole

2 moles of ions CaCl2(s) H2O > Ca2+(aq) + 2 Cl-(aq)

3 moles of ions Task 8k

1. Write the equation for the dissolution of aluminum sulfate in water.

2. How many moles of aluminum ions and sulfate ions are produced by dissolving 1 mole of aluminum sulfate?

3. What is the total number of moles of ions produced by dissolving 1 mole of aluminum

sulfate?


Precipitation Reactions Although no ionic compound is completely insoluble, compounds of very low solubility can be considered insoluble for most practical purposes. There are some general rules to help predict whether an ionic compound made of certain ions is soluble. It is important that you learn these.

SOLUBLE COMPOUNDS INSOLUBLE COMPOUNDS

Group 1 and ammonium Hydroxides (EXCEPT Group 1 and Ammonium. Hydroxides of Ca2+, Sr2+, and Ba2+

are slightly soluble) Nitrates, Hydrogen carbonates, Chlorates, Perchlorates, and Acetates

Chlorides, Bromides, and Iodides (EXCEPT those of Pb2+, Ag+, and Hg2

2+) Carbonates, Phosphates, Chromates, Silicates and Sulfides

(EXCEPT Group 1 and Ammonium. Sulfides of Group 2 are soluble) Sulfates

(EXCEPT Ca2+, Sr2+, Ba2+, Ag+, Pb2+, and Hg22+ )

This information is useful in predicting what happen if solutions of two different soluble compounds are mixed. If the mixing results in a combination of ions that forms an insoluble compound, a double-displacement reaction and precipitation will occur. Precipitation occurs when the attraction between the ions is greater than the attraction between the ions and surrounding water molecules. A precipitate is a solid that is produced as a result of a chemical reaction in aqueous solution. EXAMPLE: Will a precipitate form when solutions of ammonium sulfide and cadmium nitrate are combined? Using the solubility rules: Ammonium sulfide is soluble. Cadmium nitrate is soluble. Ammonium nitrate is soluble. Cadmium sulfide is insoluble. Therefore cadmium sulfide will precipitate (ppt).


Net ionic equations Reactions in aqueous solutions are usually represented as net ionic equations rather than formula equations. A net ionic equation includes only those compounds and ions that undergo a chemical change in a reaction in an aqueous solution. To write a net ionic equation:

1. Write a chemical equation. 2. Write an overall ionic equation. 3. Remove ions that are spectator ions. 4. Write the net ionic equation. 5. Balance elements and charges.

Using the example from the previous page: Solutions of ammonium sulfide and cadmium nitrate are mixed. 1. Write a chemical equation. Use solubility rules to determine physical state.

(NH4)2S(aq) + Cd(NO3)2(aq) Æ NH4NO3(aq) + CdS(s)

2. Write an overall ionic equation. Ionic compounds that are soluble are written as ions. (Strong acids and bases are also written in ionic form.)

NH4

+(aq) + S2-(aq) + Cd2+(aq) + NO3-(aq) Æ NH4

+(aq) + NO3-(aq) + CdS(s)

Notice that I do not include the number of ions here. I usually wait until I balance at the end. 3. Remove ions that are spectator ions. Ions that do not take part in a chemical reaction and are forum in solution both before and after the reaction are spectator ions. Their formulas and charges must match exactly.

NH4+(aq) + S2-(aq) + Cd2+(aq) + NO3

-(aq) Æ NH4+(aq) + NO3

-(aq) + CdS(s)

NH4+ and NO3

- are spectator ions

4. Write the net ionic equation.

S2-(aq) + Cd2+(aq) Æ CdS(s)

5. Balance elements and charges. In this case, the elements and charges are already balanced. So the final answer is:

S2-(aq) + Cd2+(aq) Æ CdS(s)

This looks like a long process, but if you remember the solubility rules, you should be able to do most of these steps in your head.


Not all reactions occur. Sometimes everything is soluble. This means that the ions are just hanging out in solution. A reaction isn’t occurring. For example: Solutions of potassium nitrate and magnesium sulfate.

KNO3(aq) + MgSO4(aq) Æ K2SO4(aq) + Mg(NO3)2(aq)

K+(aq) + NO3-(aq) + Mg2+(aq) + SO4

2-(aq) Æ K+(aq) + SO42-(aq) + Mg2+(aq) + NO3

-(aq)

Everything cancels here so there is no reaction!

Task 8l

1. Predict whether each of the following compounds is considered soluble or insoluble.

a. KCl b. NaNO3 c. AgCl d. BaSO4 e. Ca3(PO4)2 f. Pb(ClO3)2 g. (NH4)2S h. PbCl2 i. FeS j. Al2(SO4)3

2. In the following combinations, determine if a precipitate will occur. If so write a net

ionic equation. Identify the spectator ions.

a. Solutions of potassium sulfate and barium nitrate are combined.

b. Solutions of barium chloride and sodium sulfate are combined.

c. Solutions of cesium bromide and lithium iodide are combined.

d. Aqueous mercury(II) chloride is combined with aqueous potassium sulfide.

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Topic 9: Stoichiometry (Chapter 9 in Modern Chemistry p. 298)

Introduction to Stoichiometry

In this topic we will focus on the quantitative aspects of chemical reactions. That’s right; you’ll need your calculators. Composition Stoichiometry (Topic 7) deals with the mass relationships of elements in compounds. Reaction Stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Reaction stoichiometry is the subject of this topic and it is based on chemical equations and the law of conservation of mass. All reaction stoichiometry calculations must start with a balanced chemical equation. Reaction Stoichiometry Problems Problems in reaction stoichiometry will be classified by the given and the unknown. Problem Type 1: Given & unknown quantities are amounts in moles.

amount of given substance (mol)

amount of unknown substance (mol)

Problem Type 2: Given amount is in moles and unknown amount is mass in grams.



mass of unknown substance (g)

Problem Type 3: Given amount is mass in grams and unknown amount is in moles.

mass of given substance (g)



Problem Type 4: Given & unknown amounts are mass in grams. mass of given substance (g)



mass of unknown substance (g)

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Mole Ratio Solving any reaction stoichiometry problem requires the use of a mole ratio to convert from mores or grams of one substance in a reaction to moles or grams of another substance. A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. This information is obtained directly from the balanced chemical equation. For example, here is the equation for the electrolysis of melted aluminum oxide to produce aluminum and oxygen.

2 Al2O3(l) Æ 4 Al(s) + 3 O2(g)

This means: 2 moles of Al2O3 can produce 4 moles of Al and 3 moles of O2 This can be represented as a mole ratio like:

2 mol Al2O3 or 4 mol Al 4 mol Al 2 mol Al2O3

2 mol Al2O3 or 3 mol O2 3 mol O2 2 mol Al2O3

4 mol Al or 3 mol O2 3 mol O2 4 mol Al

Can you see that this would also mean: 4 moles of Al2O3 can produce 8 moles of Al and 6 moles of O2

Example Problem 1: Determine the amount in moles of aluminum that can be produced from 13.0 mol of aluminum oxide. What is the mole ratio that you will need? Al to Al2O3 The Al produce is 2x as many moles as aluminum oxide. So the answer is simply 26.0 moles of Al. It’s not always so easy to do the math in your head. If you need to you can use dimensional analysis.

13.0 mol Al2O3 x 4 mol Al = 26.0 mol Al

2 mol Al2O3

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Note: You will always use a mole ratio when you need to change from one substance in an equation to another substance in that equation. Molar Mass Remember in Topic 7 that the molar mass is the mass, in grams, of one mole of a substance. Recall that the molar mass can also be used as a conversion factor that relates the mass of a substance to the amount in moles of that same substance. You can use molar mass to go from grams to moles or to go from moles to grams. Again using the equation on p. 2:

1 mol Al2O3 = 101.96 g 1 mol Al = 26.98 g 1 mol O2 = 32.00 g These molar masses can be expressed by the following conversion factors.

101.96 g Al2O3 or 1 mol Al2O3

1 mol Al2O3 101.96 g Al2O3

26.98 g Al or 1 mol Al

1 mol Al 26.98 g Al

32.00 g O2 or 1 mol O2 1 mol O2 32.00 g O2

Example Problem 2: Determine the grams of aluminum equivalent to 26.0 moles of aluminum.

26.0 mol Al x 26.98 g Al = 701 g Al 1 mol Al

Ideal Stoichiometric Calculations Chemical equations help us make predictions about chemical reactions without having to run the reactions in the laboratory. These reactions are theoretical. They tell us the amounts of reactants and products for a given chemical reaction under ideal conditions, in which all reactants are completely converted into products. Theoretical stoichiometric calculations allow us to determine the maximum amount of product that could be obtained in a reaction. These problems are extensions of the composition stoichiometry problems that were solved in Topic 7 (mole conversions). It is important to use a logical, systematic approach to successfully solve these problems. We will use dimensional analysis.

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Conversion of Quantities in Moles (Mole-Mole stoichiometry) The Plan:

amount of amount of given substance (mol) unknown substance (mol)

To solve this plan requires one conversion factor – the mole ratio of the unknown to the given from the balanced equation.

Mole ratio

(Balanced equation)

Amount of given

substance (mol)

Amount of unknown substance

(mol)

x mol unknown = mol given

GIVEN IN THE

PROBLEM

CONVERSION FACTOR

CALCULATED

Example Problem 3 (p. 305): In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation.

CO2(g) + 2 LiOH(s) Æ Li2CO3(s) + H2O(l) How many moles of lithium hydroxide are required to react with 20 moles of CO2, the average amount exhaled by a person each day?

20 mol CO2 x 2 mol LiOH = 40 mol LiOH

1 mol CO2 Task 9a 1. Ammonia, NH3, is widely used as a fertilizer and in many household cleaners. How many moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas? 2. The decomposition of potassium chlorate is used as a source of oxygen in the laboratory. How many moles of potassium chlorate are needed to produce 15 mole of oxygen gas?

Given Conversion factor from balanced equation

Calculated answer

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Conversions of Amounts in Moles to Mass (Mole-Mass Stoichiometry) The Plan:

amount of amount of mass of given substance unknown substance unknown substance

(mol) (mol) (g) To solve this plan two conversion factors are required – the mole ratio of the unknown to the given from the balanced equation and the molar mass of the unknown.

Mole ratio

Molar mass factor

(Balanced equation)

(Periodic table)

Amount of given

substance (mol)

Mass of unknown

substance (g) x mol unknown x Molar mass of unknown (in g) =

mol given 1 mol of unknown

GIVEN IN

THE PROBLEM

CONVERSION FACTORS

CALCULATED

Example Problem 4 (p. 306): In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide? First, write a balanced equation

6 CO2(g) + 6 H2O(l) Æ C6H12O6(s) + 6 O2(g)

3.00 mol H2O x 1 mol C6H12O6 x

180.18 g C6H12O6 = 90.1 g

C6H12O6 6 mol H2O 1 mol C6H12O6

Task 9b 1. When magnesium burns in air, it combines with oxygen to form magnesium oxide. What mass in grams of magnesium oxide is produced from 2.00 mol of magnesium? 2. What mass of glucose can be produced from a photosynthesis reaction that occurs using 10 mol CO2?

Given Molar mass from the periodic table

Conversion factor from balanced equation

Calculated answer

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Conversion of Mass to Amounts in Moles (Mass-Mole Stoichiometry) The Plan:

mass of amount of amount of given substance given substance unknown substance

(g) (mol) (mol) To solve this plan two conversion factors are required – the molar mass of the unknown and the mole ratio.

Molar mass factor

Mole ratio

(Periodic table)

(Balanced equation)

Mass of given

substance (g)


(mol)

x 1 mol given x mol unknown = Molar mass of given (in g) mol given

GIVEN IN

THE PROBLEM

CONVERSION FACTORS

CALCULATED

Example Problem 5 (p. 309): The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.

NH3(g) + O2(g) Æ NO(g) + H2O(g) The reaction is run using 824 g ammonia and excess oxygen. How many moles of H2O are formed? Even though we were given the equation, it is not balanced. Balance the equation.

4 NH3(g) + 5 O2(g) Æ 4 NO(g) + 6 H2O(g)

824 g NH3 x 1 mol NH3 x 6 mol H2O = 72.5 mol H2O 17.04 g NH3 4 mol NH3

Task 9c 1. How many moles or mercury(II) oxide, HgO, is required to decompose to produce 124 g of oxygen?

Given Conversion factor from balanced equation Molar mass from the

periodic table

Calculated answer

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Mass-Mass Calculations Mass-mass calculations are more practical than other mole calculations. You can never measure moles directly but you can measure mass directly. Mass-mass calculations can be viewed as the combination of the other types of problems. The Plan:

mass of amount of amount of amount of given substance given substance unknown substance unknown substance

(g) (mol) (mol) (g) To solve this plan two conversion factors are required – the molar mass of the unknown and the mole ratio.

Molar mass factor

Mole ratio

Molar mass factor

(Periodic table)

(Balanced equation)

(Periodic table)

Mass of given

substance (g)


(mol)

x 1 mol given x

mol unknown x

Molar mass of unknown (in g)

= Molar mass of given

(in g) mol given 1 mol unknown

GIVEN IN THE

PROBLEM CONVERSION FACTORS

CALCULATED

Example Problem 6 (p. 310): Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation.

Sn(s) + 2 HF(g) Æ SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?

30.00 g HF x 1 mol HF x 1 mol SnF2 x 156.71 g SnF2

= 117.5 g SnF2 20.00 g HF 2 mol HF 1 mol SnF2

Given

Molar mass from the periodic table


Conversion factor from balanced equation

Calculated answer

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Task 9d 1. When copper metal is added to silver nitrate in solution, silver metal and copper(II) nitrate are produced. What mass of silver is produced from 100.0 g Cu? 2. What mass of aluminum is produced by the decomposition of 5.0 kg Al2O3? Other conversion factors Since all of these stoichiometric calculations must go through a mole ratio step, any conversion factor that can be used to change one to another can be use in a problem. For example: molar volume of a gas 22.4 L of gas at STP mole of gas Example Problem 7: 32. 7 g of potassium chlorate decomposes to produce potassium chloride and oxygen. What volume of oxygen gas will be produced?

2 KClO3(s) Æ 2 KCl(s) + 3 O2(g)

32.7 g KClO3 x 1 mol KClO3 x 1 mol O2 x 22.4 L O2

= 5.97 L O2 122.6 g KClO3 1 mol KClO3 1 mol O2

Limiting Reactants and Percentage Yield

Limiting Reactants In the laboratory, a reaction is rarely carried out with exactly the required amount of each of the reactants. In many cases, one or more reactants is present in excess; that is, there is more than the exact amount required to react. Once one of the reactants is used up, no more products can be formed. The substance that is completely used up first in a reaction is called the limiting reactant. The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The substance that is not used up completely in a reaction is called the excess reactant. A limiting reactant may also be referred to as a limiting reagent.

Molar volume of a gas at STP


Conversion factor from balanced

equation

Given Calculated answer

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If I want to make 3 batches of chocolate chip cookies, but only have enough sugar to make 1 batch, then I can only make 1 batch. The sugar is the limiting reactant. The other ingredients are excess reactants. Example Problem 8 (p. 313): Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation.

SiO2(s) + 4 HF(g) Æ SiF4(g) + 2 H2O(l) If 6.0 mol HF is added to 4.5 mol SiO2, which one is the limiting reactant?

There are 2 ways you can approach this problem. Method 1: Determine the amount of product that each of these reactants will produce. The reactant that produces the least amount of product is the limiting reactant.

6.0 mol HF 1 mol SiF4 = 1.5 mol SiF4 produced

4 mol HF

4.5 mol SiO2 1 mol SiF4 = 4.5 mol SiF4 produced

1 mol SiO2

Since HF produced the least amount of product, it is the limiting reactant.

Method 2: Compare the two reactants. Determine from that information which is limiting. 6.0 mol HF 1 mol SiF2 = 1.5 mol SiF2

produced 4 mol HF

Since SiF2 is in excess, HF is limiting.

From here you could find out the mass of SiF4 or H2O produced. You could also find out how much SiF2 is leftover after the limiting reactant is used up.

This is the lowest # of moles so this reaction stopped first.

6.0 mol of HF needs 1.5 mol of SiF2. The problem gives 4.5 mol of SiF2 so there is plenty of SiF2.


Example Problem 9 (p. 314): The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. This substance can also be made in the laboratory by the reaction between red-hot iron and steam according to the following equation.

3 Fe(s) + 4 H2O(g) Æ Fe3O4(s) + 4 H2(g)

When 36.0 g H2O is mixed with 67.0 g Fe, what mass in grams of black iron oxide is produced? What reactant is limiting? How much of the excess reactant is left over? Since we do not know what the limiting reactant is, let’s find that first.

36.0 g H2O 1 mol H2O 3 mol Fe 55.8 g Fe = 83.7 g Fe

18.0 g H2O 4 mol H2O 1 mol Fe

This means that 36.0 g of H2O need 83.7 g of Fe. The problem only provided 67.0 g of Fe therefore Fe is the limiting reactant. Now that we know that Fe is the limiting reactant, we will start the rest of the problems with the 67.0 g of Fe. Calculation of the mass of Fe3O4

67.0 g Fe 1 mol Fe 1 mol Fe3O4 231 g Fe3O4 = 92.5 g Fe3O4

55.8 g Fe 3 mol Fe 1 mol Fe3O4

Now we need to find how much H2O is used up by the 67.0 g of Fe. Then we can subtract from 36.0 g that was given in the problem.

67.0 g Fe 1 mol Fe 4 mol H2O 18.0 g H2O = 28.8 g H2O

55.8 g Fe 3 mol Fe 1 mol H2O

This means that 67.0 g of Fe reacts with 28.8 g of H2O. The problem gave us 36.0 g of H2O. We need to subtract. 36.0 g H2O given – 28.8 g needed = 7.2 g H2O left over.


Task 9e 1. Zinc & sulfur react to form zinc sulfide according to the following equation.

8 Zn(s) + S8(s) Æ 8 ZnS(s) a. If 2.00 mol of Zn are heated with 1.00 mol of S8, identify the limiting reactant. b. How many moles of excess reactant remain? c. How many moles of the product are formed? 2. 90.0 g of FeCl3 reacts with 52.0 g of H2S. What is the limiting reactant? What is the mass of HCl produced? What mass of excess reactant remains after the reaction? Percentage Yield So far the amounts of products calculated represent theoretical yields. The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. This would be the amount produced if everything went exactly as the balanced chemical equation is written, with absolutely no waste. Of course, in most chemical reactions, the amount of product obtained is less than the theoretical yield. There may be impurities in the reactants, or byproducts could be produced in competing side reactions. Also many reactions do not go to completion. This measured amount of a product obtained from a reaction is called the actual yield of that product. Chemists are interested in the efficiency of a reaction. The efficiency is expressed by comparing the actual and theoretical yields. The percentage yield is the ration of the actual yield to the theoretical yield, multiplied by 100.

percentage yield = actual yield x 100 theoretical yield


Example Problem 10 (p. 317): Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine, as represented by the following equation.

C6H6(l) + Cl2(g) Æ C6H5Cl(l) + HCl(g)

When 36.8 g C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percentage yield of C6H5Cl? You have two givens in this problem: 36.8 g of C6H6 (a reactant) and 38.8 g of C6H5Cl (a product). You can use the 36.8 g of C6H6 to determine the theoretical yield of the problem.

36.8 g C6H6 x 1 mol C6H6 x 1 mol C6H5Cl x 112.56 g C6H5Cl = 53.0 g C6H5Cl 78.12 g C6H6 1 mol C6H6 1 mol C6H5Cl

This is the

theoretical yield

Now that we have the theoretical yield, we can use the percentage yield formula.

percentage yield = actual yield x 100 theoretical yield

percentage yield = 38.8 g x 100 = 73.2%

53.0

Percent Yield Task 9f 1. Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst.

CO(g) + 2 H2(g) CH3OH(l)

If 75.0 g of CO reacts to produce 68.4 g CH3OH, what is the percentage yield of CH3OH?


2. Aluminum reacts with excess copper(II) sulfate according to the reaction given below. If 1.85 g of Al react and the percentage yield of Cu is 56.6%, what mass of Cu is produced?

Al(s) + CuSO4(aq) Æ Al2(SO4)3(aq) + Cu(s) (unbalanced)

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